ML Aggarwal Maths for Class 12 Solutions Pdf – Understanding ISC Mathematics Class 12 Solutions

ISC Mathematics Class 12 Solutions – ML Aggarwal Class 12 ISC Solutions

Section A

ML Aggarwal Class 12 Solutions ISC Pdf Chapter 1 Relations and Functions

ML Aggarwal Class 12 Solutions Chapter 2 Inverse Trigonometric Functions

ML Aggarwal Maths for Class 12 Solutions Pdf Chapter 3 Matrices

Understanding ISC Mathematics Class 12 Solutions Chapter 4 Determinants

ML Aggarwal Class 12 ISC Solutions Chapter 5 Continuity and Differentiability

ML Aggarwal Class 12 ISC Solutions Chapter 6 Indeterminate Forms

Class 12 ML Aggarwal Solutions Chapter 7 Applications of Derivatives

ISC Mathematics Class 12 Solutions Chapter 8 Integrals

ISC Maths Class 12 Solutions Chapter 9 Differential Equations

Class 12 ISC Maths Solutions Chapter 10 Probability

Section B

ISC Class 12 Maths Solutions Chapter 1 Vectors

ISC Class 12th Maths Solutions Chapter 2 Three Dimensional Geometry

Maths Class 12 ISC Solutions Chapter 3 Applications of Integrals

Section C

ISC 12th Maths Solutions Chapter 1 Application of Calculus in Commerce and Economics

Maths ISC Class 12 Solutions Chapter 2 Linear Regression

ISC Class 12 Mathematics Solutions Chapter 3 Linear Programming

Case Study Based Questions

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction MCQs

Regular engagement with Understanding ISC Mathematics Class 11 Solutions Chapter 4 Principle of Mathematical Induction MCQs can boost students’ confidence in the subject.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction MCQs

Choose the correct answer from the given four options in questions (1 to 11) :

Question 1.
Let P (n) : 2n < ∠n then the smallest positive integer for which P (n) is true is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Given P (n) = 2n < ∠n
For n = 1 ;
2 < ∠1 = 1, which is false
For n = 2 ;
22 < ∠2 2, which is false
For n = 3 ;
23 < ∠3 = 6, which is false
When n = 4 ;
24 = 16 < ∠4 = 24, which is true

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction MCQs

Question 2.
Consider the statement P (n) : n2 – n + 41 is prime. Then which of the following is true ?
(a) Both P (3) and P (5) are true.
(b) P (3) is true but P (5) is false.
(c) Both P (3) and P (5) are false.
(d) P (3) is false but P (5) is true.
Answer:
(a) Both P (3) and P (5) are true.

Given P(n) : n3 – n + 41 is prime
Here p (3) ;
32 – 3 + 41 = 47 which is prime
∴ P (3) is true.
and P (5) ;
52 – 5 + 41 = 61, which is prime
∴ P (5) is true.

Question 3.
Let P(n) = 1 + 3 + 5 + …………………. + (2n – 1) = 3 + n2, then which of the following is true?
(a) P (3) is correct
(b) P (2) is correct
(c) P (m) ⇒ P (m + 1)
(d) P (m) ⇏ P (m + 1)
Answer:
(c) P (m) ⇒ P (m + 1)

Given P (n) = 1 + 3 + 5 + ………………….. + (2n – 1) = 3 + n2
P (3) = 1 + 3 + 5 = 9
and 3 + n2 = 3 + 32 = 12
∴ P (3) is not correct.
Here P (2) = 1 + 3 ≠ 3 + 2
∴ P (2) is not correct.
Let P (m) is true
i.e. 1 + 3 + 5 + ……………………. + 2m – 1 = 3 + m2
Here P(m + 1) : 1 + 3 + 5 + …………………… + (2m – 1) + (2m + 1) = 3 + m2 + 2m + 1 = m2 + 2m + 4
= 3 + (m + 1)2
∴ P (m + 1) is true.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction MCQs

Question 4.
Let P (n) : n2 < 2n, n > 1, then the smallest positive integer for which P (n) is true ?
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(d) 5

Given P (n) : n2 < 2n
P (2) : 22 < 22 which is false
P (3) : 32 <23, which is false
P (4) : 42 < 24, which is false
P (5) : 52 = 25 < 32 = 25 which is true.

Question 5.
Consider the statement P (n): iOn + 3 is prime, then which of the following is not true?
(a) P (1)
(b) P (2)
(c) P (3)
(d) P (4)
Answer:
(c) P (3)

Given P (n) : 10n + 3 is prime
Hence P (1) = 10 + 3 = 13 is prime
P (2) : 10 × 2 + 3 = 13 is prime
P (3) : 10 × 3 + 3 = 33 which is not prime.

Question 6.
Let P (n) be the statement n (n + 1) (n + 2) is an integral multiple of 12, then which of the following is not true?
(a) P (3)
(b) P (4)
(c) P (5)
(d) P (6)
Answer:
(c) P (5)

Let P (n) : n (n + 1) (n + 2) is an integral multiple of 12.
Here P (3) : 3 (3 + 1) (3 + 2) = 60 which is an integral multiple of 12.
P (4) : 4 (4 + 1) (4 + 2) = 120, which is an integral multiple of 12.
P (5) : 5 (5 + 1) (5 + 2) = 5 × 6 × 7 = 210, which is not an integral multiple of 12.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction MCQs

Question 7.
Let P (n) : (2n + 1) < 2n. then the smallest positive integer for which P (n) is true ?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(b) 3

Given P (n) : (2n + 1) < 2n
P (2) : 2 × 2 1 = 5 > 22 = 4
∴ P(2) is false.
P (3) : 2 × 3 + 1 = 7 < 2 = 8
∴ P (3) is true.

Question 8.
If 10n + 3 . 4n + 2 + k is divisible by 9, ∀ n ∈ N. then the least positive integral value of k is
(a) 1
(b) 3
(c) 5
(d) 7
Answer:
(c) 5

Let P (n) : 10n + 3 . 4n + 2 + k is divisible by 9.
∴ P (1): 10 + 3 × 43 + k i.e. 202 + k is divisible by 9
∴ k = 5
∵ 207 is divisible by 9. .

Question 9.
For all n ∈ N, 3 . 52n + 1 + 23n + 1 is divisible by
(a) 17
(b) 19
(c) 23
(d) 25
Answer:
(a) 17

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction MCQs

Let P (n) : 3 . 52n + 1 + 23n + 1 is divisible by same number.
P (1) ; 3 × 52 + 1 + 23 + 1 = 375 + 16 = 391 which is divisibel by 17

Question 10.
If xn – 1 is divisible by x – k, then the least positive integral value of k is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Since xn – 1 is divisible by x – k
∴ xn – 1 = f (x) (x – k)
⇒ f (x) = \(\frac{x^n-1}{x-k}\)
Since f (x) is a polynomial in x
∴ k must be equal to 1.
Thus,
\(\frac{x^n-1}{x-1}=\frac{(x-1)\left(x^{n-1}+x^{n-2}+\ldots .+1\right)}{x-1}\)
= xn – 1 + xn – 2 + ………………………… + 1

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction MCQs

Question 11.
A student was asked to prove a statement P (n) by induction. 11e proved that P (k + 1) is true whenever P (k) ¡s true for all k ≥ 5 ∈ N and also that P (5) is true on the basis of this he conclude that P (n) is true
(a) ∀ n ∈ W
(b) ∀ n > 5
(c) ∀ n ≥ 5
(d) ∀ n < 5
Solution:
(c) ∀ n ≥ 5

Since P (5) is true.
∴ P (n) is true for n 5 and P (k) is given to be true for all k ≥ 5 ∈ N and student proved that P (k + 1) is true.
Thus by Mathematical induction, P (n) is true for all n ≥ 5.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Peer review of ML Aggarwal Maths for Class 11 Solutions Chapter 4 Principle of Mathematical Induction Ex 4.2 can encourage collaborative learning.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Using the principle of mathematical induction prove that (1 to 22) for all n ∈ N :

Question 1.
2 + 4 + 6 + …………… + 2n = n2 + n.
Solution:
Let P (n) be the statement :
2 + 4 + 6+ ………………….. + 2n = n2 + n
Now P (1) means, 2 = 12 + 1 = 2, which is true
= P(1) is true.
Let P (m) be true
i.e. 2 + 4 + 6 + ……………….. + 2m = m2 + m
For P (m + 1) ;
2 + 4 + 6 …………………. + 2 (m + 1) = 2 + 4 + 6 + ………….. + 2m + 2m + 2
= m2 + m + 2m + 2 [using (1)]
= m2 + 3m + 2
= (m + 1) (m + 2)
= (m + 1) [m + 1 + 1]
= (m + 1)2 + (m + 1)
Thus, P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 2.
1 + 4 + 7 + ……………………. + (3n – 2) = \(\frac{1}{2}\) n (3n – 1)
Solution:
Let P (n) be the statement :
1 + 4 + 7 + ……………………. + (3n – 2) = \(\frac{1}{2}\) n (3n – 1)
Now P (1) means,
1 = \(\frac{1}{2}\) × (3 – 1)
= \(\frac{1}{2}\) × 2 = 1, which is true
⇒ P(1) is true.
Let P (m) is true i.e. 1 + 4 + 7 + ……………………….. + 3m – 2 = \(\frac{1}{2}\) m (3m – 1)
For P (m + 1) :
1 + 4 + 7 …………… + 3m – 2 = \(\frac{1}{2}\) m (3m – 1)
= \(\frac{1}{2}\) m (3m – 1) + 3m + 1
= \(\frac{1}{2}\) [3m2 – m + 6m + 2]
= \(\frac{1}{2}\) [3m2 + 5m + 2]
= \(\frac{1}{2}\) (m + 1) (3m + 2)
= \(\frac{1}{2}\) (m + 1) [3 (m + 1) – 1]
Thus P(m + 1) is true.
Hence, by mathematical induction, P (n) is true for all n ∈ N.

Question 3.
3x+ 6x+ 9x + …………………………. to n terms = \(\frac{3}{2}\) n (n + 1) x.
Solution:
Let P (n) be the statemen t:
3x + 6x + 9x …………………….. + n terms = \(\frac{3}{2}\) n (n + 1) x
Now P (1) means,
3x = \(\frac{3}{2}\) 1 (1 + 1)x
= \(\frac{3}{2}\) × 2x = 3x, which is true
∴ P (1) is true.
Let P (m) is true i.e. 3x + 6x + 9x + ………………….. + m terms = \(\frac{3}{2}\) m (m + 1) x ……………….(1)
For P (m + 1) :
3x + 6x + 9x + ……………… + (m + 1) terms = 3x + 6x + 9x + ………….. + 3mx + 3 (m + 1) x
= \(\frac{3}{2}\) m (m + 1) x + 3 (m + 1) x [using (1)]
= \(\frac{3(m+1) x}{2}\) [m + 2]
= \(\frac{3}{2}\) (m + 1) (m + 1 + 1) x
⇒ P(m + 1) is true.
Thus by mathematical induction, P (n) is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 4.
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^n}=1-\frac{1}{2^n}\).
Solution:
Let P (n) be the statement:
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^n}=1-\frac{1}{2^n}\)
Now P (1) means,
\(\frac{1}{2}=1-\frac{1}{2^1}=1-\frac{1}{2}=\frac{1}{2}\) which is true
⇒ P (1) is true.
Let P (m) is true.
i.e. \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^m}=1-\frac{1}{2^m}\) ………………..(1)
Now P (m + 1) ;
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots .+\frac{1}{2^m}+\frac{1}{2^{m+1}}\)
= \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \ldots .+\frac{1}{2^m}\right)+\frac{1}{2^{m+1}}\)
= 1 – \(\frac{1}{2^m}+\frac{1}{2^{m+1}}\) [using (1)]
= 1 – \(\frac{1}{2^m}\left[1-\frac{1}{2}\right]\)
= 1 – \(\frac{1}{2^m} \times \frac{1}{2}\)
= 1 \(\frac{1}{2^{m+1}}\)
⇒ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 5.
12 + 32 + 52 + ……………… to n terms = \(\frac{n\left(4 n^2-1\right)}{3}\)
Solution:
Let p (n) be the statement :
12 + 32 + 52 + ……………… to n terms = \(\frac{n\left(4 n^2-1\right)}{3}\)
Now P (1) means 12 = 1, which is true
∴ P (1) is true.
Let P (m) be true.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2 1

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 6.
13 + 23 + ……………… + n3 = \(\left(\frac{n(n+1)}{2}\right)^2\)
Solution:
Hence P(n) be the statement :
13 + 23 + ……………… + n3 = \(\left(\frac{n(n+1)}{2}\right)^2\)
For n = 1 ;
P (1) means n3 = \(\left(\frac{1(1+1)}{2}\right)^2\) = 1, which is true.
∴ P(1) is true
Let P (n) be true for n = m
Thus,
13 + 23 + ……………… + m3 = \(\left(\frac{m(m+1)}{2}\right)^2\) ………………..(1)
For P (m + 1) :
13 + 23 + …………………. + m3 + (m + 1)3 = {13 + 23 + ………………….. + m3} + (m + 1)3
= \(\left[\frac{m(m+1)}{2}\right]^2\) + (m + 1)3 [using (1))
= \(\frac{(m+1)^2}{4}\) [m2 + 4m + 4]
= \(\frac{(m+1)^2(m+2)^2}{4}\)
= \(\left[\frac{(m+1)(\overline{m+1}+1)}{2}\right]^2\)
Thus P(n) is true for n = m + 1.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 7.
3 . 6 + 6 . 9 + 9. 12 + ……………………… + 3n (3n + 3) = 3n (n + 1) (n + 2).
Solution:
Let P (n) be the statement :
3 . 6 + 6 . 9 + 9. 12 + ……………………… + 3n (3n + 3) = 3n (n + 1) (n + 2)
Now P (1) means,
3.6 = 3 × 1 (1 + 1) (1 + 2)
⇒ 18 = 3 × 2 × 3 = 18, which is true
∴ P (1) is true.
Let P (m) is true
i.e. 3 . 6 + 6 . 9 + 9 . 12 + …………………….. + 3m (3m + 3) = 3m (m + 1) (m + 2)
For P (m + 1) :
3 . 6 + 6 . 9 + 9 . 12 + …………………….. + 3m (3m + 3) + 3m (m + 1) (m + 2) = 3m (m + 1) (m + 2) + 3 (m + 1) (3m + 6) [using (1)]
= 3 (m + 1) [m (m + 2) + 3m + 6]
= 3 (m +1) [m2 + 5m + 6]
= 3 (m + 1) (m + 2) (m + 3)
= 3 (m + 1) (m + 1 + 1) (m + 1 + 2)
⇒ P (m + 1)is true.
Hence by induction P(n) is truc for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 8.
1 . 2 . 3 + 2 . 3 . 4 + …………………… + n (n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
Solution:
Let P (n) be the statement :
1 . 2 . 3 + 2 . 3 . 4 + …………………… + n (n + 1) (n + 2) = \(\frac{n(n+1)(n+2)(n+3)}{4}\)
For n = 1 ;
P (1) = 1 . 2 . 3 = 6
= \(\frac{1(1+1)(1+2)(1+3)}{4}\)
= \(\frac{24}{4}\) = 6
Thus P (1) is true.
Let us assume that P (n) is true for n = m.
∴ 1 . 2 . 3 + 2 . 3 . 4 + …………………… + m (m + 1) (m + 2) = \(\frac{m(m+1)(m+2)(m+3)}{4}\) …………………..(1)
For P (m + 1) ;
1 . 2 . 3 + 2 . 3 . 4 + …………………… + m (m + 1) (m + 2) + (m + 1) (m + 2) (m + 3) = \(\frac{m(m+1)(m+2)(m+3)}{4}\) + (m + 1) (m + 2) (m + 3)
= \(\frac{(m+1)(m+2)(m+3)}{4}\) [m + 4]
= \(\frac{(m+1)(\overline{m+1}+1)(\overline{m+1}+2)(\overline{m+1}+3)}{4}\)
∴ P (n) is true forn=m+I
Hence by mathematical induction result is true V n E N.

Question 9.
1 . 2 + 2 . 22 + 3 . 23 + ………………. + n . 2n = (n – 1) . 2n+1 + 2.
Solution:
Let P (n) be the statement:
1 . 2 + 2 . 22 + 3 . 23 + ………………… + n . 2n (n – 1) 2n+1 + 2
Here P (1) :
1 . 2 = 2 = (1 – 1) 21+1 + 2 = 2, which is true
∴ P (1) is true.
Let P(n) is true for n = m
i.e. 1 . 2 + 2 . 22 + ………………… + m . 2m = (m – 1) 2m+1 + 2 ………………..(1)
For P (m + 1) :
1 . 2 + 2 . 22 + ………………….. + m . 2m + (m + 1) 2m + 1
= (m – 1) 2m + 1 + (m + 1) 2m+1 + 2 [using (1)]
= 2m+1 (m – 1 + m ÷ I) + 2
= m . 2m + 2 + 2
∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 10.
\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)\) = n + 1
Solution:
Let P (n) be the statement :
\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \ldots\left(1+\frac{1}{n}\right)\) = n + 1
Now P (1) means,
1 + \(\frac{1}{1}\) = 1 + 1, which is true
∴ P (1) is true.
Let P (m) is true.
i.e. \(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right) \ldots\left(1+\frac{1}{m}\right)\) = m + 1 …………………(1)
For P (m + 1) ;
\(\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right) \ldots\left(1+\frac{1}{m}\right)\left(1+\frac{1}{m+1}\right)=(m+1)\left[1+\frac{1}{m+1}\right]\)
= m + 1 + 1
⇒ P (m + 1) is true.
Thus by mathematical induction, result P (n) is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 11.
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \cdots\left(1-\frac{1}{n+1}\right)=\frac{1}{n+1}\).
Solution:
Let P (n) be the statement :
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \cdots\left(1-\frac{1}{n+1}\right)=\frac{1}{n+1}\)
Now P (1) means,
\(1-\frac{1}{2}=\frac{1}{1+1}\)
i.e. \(\frac{1}{2}=\frac{1}{2}\), which is true.
⇒ P (1) is true.
Let P (m) is true.
i.e. \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots\left(1-\frac{1}{m+1}\right)=\frac{1}{m+1}\) ………………….(1)
For P (m + 1) ;
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right) \ldots\left(1-\frac{1}{m+1}\right)\left(1-\frac{1}{m+2}\right)\)
= \(\frac{1}{m+1}\left(1-\frac{1}{m+2}\right)\)
= \(\frac{1}{m+1}\left[\frac{m+2-1}{m+2}\right]\)
= \(\frac{1}{m+2}\)
= \(\frac{1}{m+1+1}\)
⇒ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

Question 12.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\) + ……………………. to n terms = \(\frac{n}{n+1}\)
Solution:
Let P(n) be the statement :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}\) + ……………………. to n terms = \(\frac{n}{n+1}\)
i.e. \(\frac{1}{1.2}+\frac{1}{2.3}+\ldots .+\frac{1}{n(n+1)}=\frac{n}{n+1}\)
[∵ nth term of (1, 2, 3, …………………n terms) (2, 3, 4, ………………. n terms) = [1 + (n – 1) 1] [2 + (n – 1) . 1] = n (n + 1)]
Now P (1) means,
\(\frac{1}{1.2}=\frac{1}{1+1}\)
⇒ \(\frac{1}{2}=\frac{1}{2}\), which is true.
⇒ P (1) is true.
Let P (m) be true.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2 2

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 13.
\(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\ldots+\frac{1}{3 n(3 n+3)}=\frac{n}{9(n+1)}\)
Solution:
\(\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+\ldots+\frac{1}{3 n(3 n+3)}=\frac{n}{9(n+1)}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2 3

Question 14.
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\) + …………….. to n terms = \(\frac{n}{2 n+1}\).
Solution:
nth term gives series = nth term of (1, 3, 5, ………………… n terms) (nth term of 3 . 5, 7, ……………..)
= [1 + (n – 1) 2] + [3 + (n – 1) 2]
= (2n – 1) (2n + 1)
Let P (n) be the statement :

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2 4

Question 15.
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}\).
Solution:
Let P (n) be the statement :
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+\ldots+\frac{1}{(3 n-2)(3 n+1)}=\frac{n}{3 n+1}\)
Now P (1) means,
\(\frac{1}{1.4}=\frac{1}{3+1}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2 5

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 16.
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}\)
Solution:
Let P (n) be the statement :
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\ldots+\frac{1}{(2 n+1)(2 n+3)}=\frac{n}{3(2 n+3)}\)
Now P (1) means,
\(\frac{1}{3.5}=\frac{1}{3(2+3)}\)
⇒ \(\frac{1}{15}=\frac{1}{15}\), which is true
⇒ P (1) is true.
Let P (m) be true.
i.e. \(\frac{1}{3.5}+\frac{1}{5.7}+\ldots+\frac{1}{(2 m+1)(2 m+3)}=\frac{m}{3(2 m+3)}\) …………………..(1)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2 6

Question 17.
22n – 1 is divisible by 3.
Solution:
Let P (n) be the statement :
22n – 1 is divisible by 3
Now p (1) means, 22 × 1 – 1 is divisible by 3
⇒ 3 is divisible by 3, which is true
⇒ P (1) is true.
Let P (m) is true
i.e. 22m – 1 be divisible by 3
⇒ 22m – 1 = 3k for some integer k ……………….(1)
For P (m + 1) :
22 (m + 1) – 1 = 22m + 2 – 1
= 22m . 4 – 1
= (3k + 1) 4 – 1 [using (1)]
= 12k + 3
= 3 (4k + 1)
[since k ∈ 1 ∴ 4k + 1 ∈ I]
which is divisible by 3.
∴ P (m + 1) is true.
Hence by mathematical induction P (n) is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 18.
23n – 1 isdivisible by 7.
Solution:
Let P (n) be the statement:
23n – 1 is divisible by 7.
Now P (1) means,
23 × 1 – 1 is divisible by 7.
i.e. (8 – 1) i.e. 7 is divisible by 7.
⇒ P (1) is true.
Let P (m) be true
i.e. 23m – 1 is divisible by 7.
⇒ 23m – 1 = 7k for some integer k
⇒ 23m = 7k + 1 …………………..(1)
For P (m + 1) ;
23 (m + 1) – 1 = 23m + 3 – 1
= 23m – 8 – 1
= 8 (7k + 1) – 1 [using (1)]
= 56k + 7
= 7 (8k + 1).
which is divisible by 8. [∵ k ∈ I
⇒ 8k + 1 ∈ I]
P (m + 1) is true
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 19.
32n when divided by 8, leaves the remainder 1.
Solution:
We want to prove that,
32n when divided by 8, leaves the remainder I. i.e. 32n – 1 divisible by 8.
Let P (n) be the statement:
32n – 1 is divisible by 8.
Now P (1) means, 32 × 1 – 1 is divisible by 8
⇒ (9 – 1) = 8 is divisible by 8
⇒ P(1) is true.
Let P(m) be true i.e. 32m – 1 is divisible by 8
⇒ 32m – 1 = 8k for some integer k …………………(1)
For P (m + ) :
32 (m + 1) – 1 = 32m . 32 – 1
= 9 (8k + 1) – 1 [using (1)]
= 72k + 9 – 1
= 8 [9k + 1]
= 8k’ [where k’ = 9k + 1 ∈ I]
which is divisible by 8
⇒ P (m + 1) is true.
Thus by induction, P(n) is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 20.
41n – 14n is divisible by 27.
Solution:
Let P (n) be the statement :
41n – 14n is divisible by 27.
P(1) means, 411 – 141 = 27 which is divisible by 27.
∴ P(1) is true.
Let us assume that P (n) is true for n = m
i.e. 41m – 14m is divisible by 27.
⇒ 41m – 14m = 27k where k ∈ N …………………(1)
For P (m + 1) ;
41m + 1 – 14m + 1
= 41 (27k + 14m) – 14m+1
= 41 × 27k + 14m (41 – 14)
= 27 (41k + 14m)
= 27k’
where k’ = 41k + 14m ∈ N
which is divisible by 27.
Thus, P (m + 1) is true.
Hence by M.I., P (n) is true for all n ∈ N.

Question 21.
72n + 23n – 3 . 3n – 1 is divisible by 25.
Solution:
Let P (n) be the statement :
72n + 23n – 3 . 3n – 1 is divisible by 25.
Here P (1) means,
72 + 23 – 3 31 – 1 = 49 + 1 = 50 which is divisible by 25.
∴ P(1) is true.
Let us assume that P (n) is true for n = m
i.e. 72m + 23m – 3 . 3m – 1 is divisible by 25.
⇒ 72m + 23m – 3 . 3m – 1 = 25k ………………..(1)
P (m + 1) : 72 (m + 1) + 23 (m + 1) – 3 . 3m + 1 – 1
= 72 (m + 1) + 23m . 3m
= 72 . (25k – 23m – 3 . 3m – 1) + 23m 3m [using (1)]
= 49 × 25k – 23m – 3 3m – 1 – 1(49 – 8 × 3)
= 25 (49k – 23m – 3 3m – 1) = 25k’
where k’ = 49k – 23m – 3 3m – 1 ∈ N
which is clearly divisible by 25
Thus P (m + 1) is true.
Hence, by mathematical induction, P (n) is true ∀ n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 22.
7n – 3n is divisible by 4.
Solution:
Let P(n) be the statement :
7n – 3n is divisible by 4.
Now P (1) means,
71 – 31 is divisible by 4
i.e. 4 is divisible by 4 which is true
⇒ P (1) is true.
Let P (m) be true
Le. 7m – 3m is divisible by 4
⇒ 7m – 3m = 4k for some integer k ………………..(2)
For P (m + 1) ;
7m + 1 – 3m + 1 = 7m . 7 – 3m + 1
= 7 (3m + 4k) – 3m + 1 [using (1)]
= 7 . 3m + 28k – 3m . 3
= 3m (7 – 3) + 28k
= 4 [3m + 7k] = 4k’
which is divisible by 4
[where k’ = 3m + 7k ∈ I, where k ∈ I]
⇒ P(m + 1) is true.
Hence by induction, P (n) is true for all n ∈ N.

Question 23.
4n + 15n – 1 is divisible by 9.
Solution:
Let P (n) be the statement:
4n + 15n – 1 is divisible by 9
Now P (1) means,
41 + 15 – 1 = 18 is divisible by 9,
which is true
⇒ P(1) is true.
Let P (m) be true
i.e. 4m + 15m – 1 is divisible by 9
⇒ 4m + 15m – 1 = 9k for some integer k
⇒ 4m + 15m = 9k + 1 …………… (1)
For P (m + 1) :
4m + 1 + 15 (m + 1) – 1 = 4 . 4m + 15m + 14
= 4 (- 15m + 9k + 1) + 15m + 14 [using (1)]
= – 45m + 36k + 18
= 9 (- 5m + 4k + 2) = 9k’
where k’ = – 5m + 4k + 2 ∈ I
which is divisible by 9.
⇒ P(m + 1) is true.
Hence by mathematical induction, P (n)is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 24.
32n + 2 – 8n – 9 is a multiple of 64.
Solution:
Let P (n) be the statement:
32n + 2 – 8n – 9 is a multiple of 64.
Now P (1) means,
32 + 2 – 8 × 1 – 9 is multiple of 64.
i.e. 81 – 8 – 9 = 64 is a multiple of 64,
which is true
⇒ P(1) is true.
Let P (m) is true
i.e. 32m + 2 – 8m – 9 is a multiple of 64
⇒ 32m + 2 – 8m – 9 = 64k for some integer k
⇒ 32m + 2 = 8m + 9 + 64k ……………(1)
For P (m + 1) ;
32m + 4 – 8 (m + 1) – 9
= 32m + 2 . 32 – 8 (m + 1) – 9
= 9 (8m + 9 + 64k) – 8m – 17 [using(1)]
= 64m + 64k + 64
= 64 (m + k + 1)
= 64k’ where k’ = m + k + 1 ∈ I
which is divisible by 64.
⇒ P(m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

Question 25.
n (n + 1) (n + 5) is a multiple of 3.
Solution:
Let P (n) be the statement :
n (n + 1) (n + 5) is a multiple of 3.
Now P (1) means,
1 (1 + 1) (1 + 5) = 2 × 6 = 12 is a multiple of 3.
which is true
∴ P (1) is true.
Let P (m) be true i.e. m (m + 1) (m + 5) is a multiple of 3
⇒ m (m + 1) (m + 5) = 3k for some integer k
⇒ m (m2 + 8m + 5) = 3k
For P (m + 1) ;
(m + 1) (m + 2) (m + 6) = (m + 1) [m2 + 8m + 12]
= m (m2 + 8m + 12) + m2 + 8m + 12
= m (m2 + 6m + 5) + m (2m + 7) + m2 + 8m + 12
= 3k + 3m2 + 15m + 12 [using (1)]
= 3 [k + m2 + 5m + 4] = 3k’
where k’ = k + m2 + 5m + 4 ∈ N as k, m ∈ N
which is multiple of 3.
⇒ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 26.
n (n +1) (2n + 1) is divisible by 6.
Solution:
Let P (n) be the statement :
n (n + 1) (2n + 1) is divisible by 6.
Here P (1) means
1 . (1 + 1) (2 + 1) = 2 . 3 = 6 is divisible by 6, which is true,
Thus, P (1) is true.
Let us assume that P (n) is true for n = m
i.e. m (m + 1) (2m + 1) is divisible by 6.
i.e. m (m + 1) (2m + 1) = 6k, k ∈ N ………………..(1)
For P (m + 1) :
(m + 1) (m + 2) (2 (m + 1) + 1) = (m + 1) (m + 2) (2m + 3)
= m (m + 1) (2m + 1) + 2 (m + 1) (2m + 3) [using (1)]
= m (m + 1) (2m + 1) + 2m (m + 1) + 2 (m + 1) (2m + 3)
= 6k + 2 (m + 1) (m + 2m + 3) [using (1)]
= 6k + 6 (m + 1)2
= 6k’, which is divisible by 6.
where k’ = k + (m + 1)2 ∈ N
Therefore, P (m + 1) is true.
Hence by M.I, P (n) is true for all n ∈ N.

Question 27.
x2n – 1 – 1 is divisible by (x — 1), x ≠ 1.
Solution:
Let P (n) be the statement :
x2n – 1 – 1 is divisible by (x — 1), x ≠ 1
Now P (1) means, x2 × 1 – 1
i.e. x – 1 is divisible by x – 1, x ≠ 1 which is true,
Thus, P (1) is true.
Let P (m) be true
i.e. x2m – 1 – 1 is divisible by x – 1.
x2m – 1 – 1 = (x – 1) f (x)
where f (x) be polynomial in x
x2m – 1 = (x – 1) f (x) + 1 …………………(1)
For P (m + 1) ;
x2 (m + 1) – 1 – 1 = 22m + 1 – 1
= x2m – 1 + 2 – 1
= x2m – 1 x2 – 1
= [(x – 1) f (x) + 1] x2 – 1
= (x- 1) x2 f (x) + x2 – 1
= (x – 1) [x2 f (x) + x + 1]
which is divisible by (x – 1)
⇒ P (m + 1) is true.
Thus, by principal of mathematucal induction P (n) is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 28.
3n > n for all n ∈ N.
Solution”
Let P (n) be the statement :
3n > n ∀ n ∈ N
Now P (1) means,
31 = 3 > 1
∴ P (1) is true.
Let P (m) be true i.e. 3m > m ∀ m ∈ N ………………..(1)
For P (m + 1) :
3m + 1 = 3m . 3 > 3.m ≥ m + 1 ∀ m ∈ N [using (1)]
[∵ m ≥ 1 > m + m ≥ m + 1
⇒ 3m ≥ 2m + 1 > m + 1 ∀ m ∈ N]
⇒ P (m + 1) is true
Hence by mathematical induction P(n) is true for all n ∈ N.

Question 29.
1 + 2 + 3 + ……………. + n < \(\frac{1}{8}\) (2n + 1)2.
Solution:
Let P (n) be the statement :
1 + 2 + 3 + ……………. + n < \(\frac{1}{8}\) (2n + 1)2
Here P (1) means, 1 < \(\frac{9}{8}\) = \(\frac{1}{8}\) (2 . 1 + 1)2
∴ P (1) is true.
Let us assume that P (n) is true for n = m [using (1)]
∴ 1 + 2 + 3 + ………………. + m < \(\frac{1}{8}\) (2m + 1)2
Thus P (m + 1) ;
1 + 2 + 3 + ………………… + m + m + 1 < (2m + 1)2 + m + 1 [using(1)]
= \(\frac{1}{8}\) [4m2 + 4m + 1 + 8m + 8]
= \(\frac{1}{8}\) [4m2 + 12m + 9]
= \(\frac{1}{8}\) (2m + 3)2
Therefore, P (m + 1) is true.
Hence by M.I, P (n) is true for all n ∈ N.

Question 30.
5 + 55 + 555 + ………………. to n terms = \(\frac{5}{81}\) (10n + 1 – 9n – 10).
Solution:
Let P (n) be the statement :
5 + 55 + 555 + …………………. to n terms = \(\frac{5}{81}\) (10n + 1 – 9n – 10)
Here P (1) :
5 = \(\frac{5}{81}\) (100 – 9 – 10)
= \(\frac{5}{81}\) × 81 = 5, which is true.
∴ P (1) is true.
Let us assume that P (n) is true for n = m.
i.e. 5 + 55 + 555 + …………………… m terms = \(\frac{5}{81}\) (10m + 1 – 9m – 10) ………………..(1)
Here (m + 1)th term = 555 ………………. (m + 1) terms
= 5 + 5 × 10 + 5 × 102 + ……………….. (m + 1) terms
= \(\frac{5\left(10^{m+1}-1\right)}{10-1}\)
= \(\frac{5}{9}\) (10m + 1 – 1) …………………….(2)
For P (m + 1) :
5 + 55 + 555 + …………………. (m + 1) terms = (5 + 55 + …………………. m terms) + (m + 1)th term
= \(\frac{5}{81}\) (10m + 1 – 9m – 10) + \(\frac{5}{9}\) (10m + 1 – 1) [using (1) and (2)]
= \(\frac{5}{81}\) [10m + 1 – 9m – 10 + 9 . 10m + 1 – 9]
= \(\frac{5}{81}\) [(1 + 9) 10m + 1 – 9 (m + 1) – 10]
= \(\frac{5}{81}\) [10m + 2 – 9 (m + 1) – 10]
∴ P (m + 1) is true.
Hence by M.I result is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2

Question 31.
If x is not an integral multiple of π, use induction to prove that sin x + sin 3x + sin 5x + …………………. + sin (2n – 1) x = \(\frac{\sin ^2 n x}{\sin x}\). for all n ∈ N.
Solution:
Let P (n) be the statement :
sin x + sin 3x + sin 5x + …………………. + sin (2n – 1) x = \(\frac{\sin ^2 n x}{\sin x}\) ∀ n ∈ N
Now P (1) means,
sin x = \(\frac{\sin ^2 x}{\sin x}\) = sin x, which is true
⇒ P(1) is true.
Let P (m) is true
i.e. sin x + sin 3x + sin 5x + …………………. + sin (2m- 1) x = \(\frac{\sin ^2 m x}{\sin x}\) ……………….(1)
For P (m + 1) ;
[sin x + sin 3x + sin 5x + ………………………. + sin (2m – 1) x] + sin (2m + 1) x

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.2 7

⇒ P (m + 1) is true.
Hence by mathematical induction, P (n) is true for all n ∈ N.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.1

Continuous practice using ML Aggarwal Class 11 ISC Solutions Chapter 4 Principle of Mathematical Induction Ex 4.1 can lead to a stronger grasp of mathematical concepts.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.1

Question 1.
If P (n) is the statement “n (n + 1) (n + 2) is divisible by 6”, then what is P (3)?
Solution:
P (n) is the statement “n (n + 1) (n + 2) is divisible by 6″
∴ P(3) is 3 (3 + 1) (3 + 2) is divisible by 6
i.e. 60 is divisible by 6. Which is true.

Question 2.
If P (n) is the statement “10n + 3 is prime”, then show that P (1) and P (2) are true but P (3) is not true.
Solution:
Given P (n) is the statement “10n + 3 is prime”
∴ p (1) i.e. 10 × 1 + 3 is prime i.e. 13 is prime, which is true.
So P (1) is true.
p (2) i.e. 10 × 2 + 3 i.e. 23 is prime, which is true.
P (2) is true.
p (3) i.e. 10 × 3 + 3 is prime i.e. 33 is prime
which is not true since 3 and 11 are factors of 33.
Thus, P (3) is not true.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.1

Question 3.
If P (n) is the statement “n (n + 1) (n + 2) is an integral multiple of 12”, prove that P (3) and P (4) are true but P (5) is not true.
Solution:
Given P (n) is the statement “n (n + 1 )(n+2) is an integral multiple of 12”.
Now P(3) i.e. 3(3 + 1) (3 +2) is an integral multiple of 12.
∴ 60 is an integral multiple of 12, which is clearly true.
Thus, P (3) is true.
Now P (4) i.e. 4 (4 + 1) (4 + 2) is an integral multiple of 12.
∴ 4.5.6 – 120 is an integral multiple of 12
which is clearly true.
Here P (5) i.e. 5 (5 + 1) (5 + 2) is an integral multiple of 12.
∴ 5 × 6 × 7 = 210 is an integral multiple of 12 which is not true.
Thus, P (5) is not true.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.1

Question 4.
If P (n) is the statement “n2 – n + 41 is prime”, show that P (1), P (2), and P (3) are true but P (41) is not true.
Solution:
Given P (n) is the statement ‘n2 – n + 41 is prime”.
∴ P(1) is 12 – 1 + 41 is prime
⇒ 41 is prime, which ¡s clearly true.
P (2) is 22 – 2 + 41 is prime
∴ 43 is prime, which is clearly true.
P (3) is 32 – 3 + 41 is prime.
∴ 47 is prime, which is clearly true.
and P (41) is 412 – 41 + 41 is prime i.e.
(41)2 is prime
which is not true since 41 be a factor of (41)2.
Thus, P (41) is not title.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.1

Question 5.
Let P (n) is the statement “n2 + n is an even integer”. Show that if P (k) is true then P (k + 1) is also true.
Solution:
Given P (n) is the statement “n2 + n is an even integer”.
Let P (k) is true k2 + k is an even integer.
⇒ k2 + k = 2λ where λ ∈ I
Now P( k + 1) = (k + 1)2 + k + 1
⇒ k2 + 2k + 1 + k + 1
= (k2 + k) + 2k + 2
= 2λ + 2k + 2
= 2 (λ + k + 1)
= even integer
Thus P (k + 1) also true.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 4 Principle of Mathematical Induction Ex 4.1

Question 6.
If P (n) denote the statement “32n – 1 is a multiple of 8”. Show that
(i) P (1), P (2) are true
(ii) if P (m) is true then P (m + 1) is also true.
Solution:
Given P (n) denote the statement “32n – 1 is a multiple of 8”.
(i) ∴ P (1) is 32 × 1 – 1 i.e. 8 is a multiple of 8, which is true.
Thus, P (1) is true.
and P (2) is 32 × 1 – 1 = 81 – 1 = 80 is a multiple of 8
which is clearly true.

(ii) Let P (m) is true
⇒ 32m – 1 is a multiple of 8
⇒ 32m – 1 = 8λ for some integer λ
⇒ 32m = (8λ + 1) …………………(1)
Now, P (m + 1) = 32 (m + 1) – 1
= 32m . 32 – 1
= 9 (8λ + 1) – 1 [using (1)]
= 72λ + 8
= 8 (9λ + 1)
which is a multiple of 8.
Thus, P (m + 1) is true.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Effective ML Aggarwal Class 11 Solutions ISC Chapter 3 Trigonometry Chapter Test can help bridge the gap between theory and application.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 1.
Find the radian measure of an angle (internal) of a regular
(i) pentagon
(ii) hexagon
(iii) polygon of n sides.
Solution:
We know that interior angle of polygon with n sides = \(\left(\frac{2 n-4}{n}\right)\) × 90°
or \(\left(\frac{2 n-4}{n}\right) \frac{\pi}{2}\) rad
(i) In pentagon, n = 5
∴ interior angle = \(\left(\frac{2 \times 5-4}{5}\right) \frac{\pi}{2}\) rad
= \(\frac{6}{5} \times \frac{\pi}{2}\) radian
= \(\frac{3 \pi}{5}\) rad.

(ii) In case of hexagon, n = 6
∴ required interior angle = \(\left(\frac{2 \times 6-4}{6}\right) \frac{\pi}{2}\) rad
= \(\left(\frac{4}{3} \times \frac{\pi}{2}\right)\) rad
= \(\frac{2 \pi}{3}\) rad.

(iii) In case of polygon of n sides
∴ required interior angle = \(\left(\frac{2 n-4}{n}\right) \frac{\pi}{2}\)
= \(\left(\frac{n-2}{n}\right)\) π rad

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 2.
Find the value of sin (- \(\frac{41 \pi}{4}\)).
Solution:
sin (- \(\frac{4 \pi}{4}\)) = – sin (\(\frac{41 \pi}{4}\))
[∵ sin (- θ) = – sin θ]
= – sin \(\left(\frac{40 \pi+\pi}{4}\right)\)
= – sin \(\left(10 \pi+\frac{\pi}{4}\right)\)
= – sin \(\frac{\pi}{4}\)
= – \(\)
[.. sin(2nlt+0)sinOVnE 1]

Question 3.
Prove that :
(i) sin2 \(\frac{\pi}{6}\), sin2 \(\frac{\pi}{4}\), sin2 \(\frac{\pi}{3}\) are in A.P.
(ii) tan2 \(\frac{\pi}{6}\) , tan2 \(\frac{\pi}{4}\), tan2 \(\frac{\pi}{3}\) are in GP.
Solution:
(i)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 1

(ii)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 2

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 4.
Find the value of m2 sin \(\frac{1}{2}\) π – n2 sin \(\frac{3}{2}\) π + 2mn sec π.
Solution:
m2 sin \(\frac{1}{2}\) π – n2 sin \(\frac{3}{2}\) π + 2mn sec π
= m2 × 1 – n2 sin (π + \(\frac{\pi}{2}\)) + 2mn \(\frac{1}{\cos \pi}\)
= m2 – n2 (9- 1) + 2mn (- 1)
= m2 + n2 – 2mn
= (m – n)2

Question 5.
What is the maximum value of 3 – 7 cos 5x?
Solution:
Since – 1 ≤ cos 5 x ≤ 1
⇒ + 7 ≥ – 7 cos 5x ≥ – 7
⇒ – 7 ≤ – 7 cos 5x ≤ 7
⇒ 3 – 7 ≤ 3 – 7 cos 5x ≤ 3 + 7
⇒ – 4 ≤ 3 – 7 cos 5x ≤ 10
Thus, max. value of 3 – 7 cos 5x = 10

Question 6.
What is the minimum value of 4 + 5 sin (3x – 2) ?
Solution:
Since – 1 ≤ sin (3x – 2) ≤ 1
[∵ |sin θ| ≤ 1 ∀ θ]
⇒ – 5 ≤ 5 sin (3x – 2) ≤ 5
⇒ 4 – 5 ≤ 4 + 5 sin (3x – 2) ≤ 4 + 5
⇒ – 1 ≤ 4 + 5 sin (3x – 2) ≤ 9
Thus, min. value of 4 + 5 sin (3x – 2) = – 1.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 7.
What is the maximum value of sin x cos x?
Solution:
Let f(x) = sin x cos x
= \(\frac{1}{2}\) (2 sin x cos x)
= \(\frac{\sin 2 x}{2}\)
since – 1 ≤ sin 2x ≤ 1
⇒ – \(\frac{1}{2}\) ≤ \(\frac{1}{2}\) sin 2x ≤ \(\frac{1}{2}\)
⇒ – \(\frac{1}{2}\) ≤ f(x) ≤ \(\frac{1}{2}\)
Thus, max. value of f (x) = \(\frac{1}{2}\).

Question 8.
What is the maximum value of 3 sin x – 4 sin3 x?
Solution:
Let f (x) = 3 sin x – 4 sin3 x
= sin 3x
since – 1 ≤ sin 3x ≤ 1
Thus, max. value of f (x) = 1.

Question 9.
What is the minimum value of 3 cos x – 4 cos3 x?
Solution:
Let f (x) = 3 cos x – 4 cos3 x
= – cos 3x
since, – 1 ≤ cos 3x ≤ 1
⇒ 1 ≥ – cos3x ≥ – 1
⇒ – 1 ≤ — cos 3x ≤ 1
⇒ – 1 ≤ f (x) ≤ 1
Thus, min value of f (x) = – 1.

Question 10.
What is the least value of 2 sin2 x + 3 cos2 x?
Solution:
Let f (x) = 2 sin2 x + 3 cos2 x
= 2 (sin2 x + cos2x) + cos2 x
= 2 + cos2 x
since cos2 x ≥ 0
⇒ 2 + cos2 x ≥ 2
⇒ f (x) ≥ 2
Thus, least value of f (x) = 2.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 11.
What is the maximum value of sin x + cos x?
Solution:
Let f (x) = sin x + cos x

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 3

Question 12.
What is the minimum value of sin x – cos x?
Solution:
Let f (x) = sin x – cos x
= √2 (\(\frac{1}{\sqrt{2}}\) sin x – \(\frac{1}{\sqrt{2}}\) cos x)
= √2 [cos \(\frac{\pi}{4}\) sin x – sin \(\frac{\pi}{4}\) cos x]
= √2 sin (x – \(\frac{\pi}{4}\))
since – 1 ≤ sin (x- \(\frac{\pi}{4}\)) ≤ 1
⇒ – √2 ≤ √2 sin (x – \(\frac{\pi}{4}\)) ≤ √2
⇒ – √2 ≤ f (x) ≤ √2
Thus, min value of f (x) = – √2

Question 13.
If sin 2x = cos 3x and 0 ≤ x < \(\frac{\pi}{2}\), then find the value of x.
Solution:
Given sin 2x = cos 3x

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 4

Question 14.
A railway carriage is travelling along a circular railway track of radius 1500 metres with a speed of 66 km / hour. Find the angle in degrees turned by the engine in 10 seconds.
Solution:
Given radius of circular railway track = r
= 1500 cm
distance covered by a railway carriage in 1 hour = 66 km
So distance covered by a railway carriage in 1 second = \(\frac{66 \times 1000}{60 \times 60}\)
Thus distance covered by a railway carriage in 10 seconds
s = \(\frac{10 \times 60 \times 1000}{60 \times 60}\) m
= \(\frac{550}{3}\) m
∴ required angle turned by engine in 10 seconds θ = \(\frac{s}{r}\)
∴ θ = \(\frac{\frac{550}{3}}{1500}\)
= \(\frac{550}{3 \times 1500}=\frac{11}{90}\) rad
= \(\frac{11}{90} \times \frac{180^{\circ}}{22}\) × 7
[∵ π rad = 180°]
= 7°

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 15.
If tan x = \(\frac{a}{b}\), show that \(\frac{a \sin x-b \cos x}{a \sin x+b \cos x}=\frac{a^2-b^2}{a^2+b^2}\).
Solution:
Given tan x = \(\frac{a}{b}\)
L.H.S. = \(\frac{a \sin x-b \cos x}{a \sin x+b \cos x}\)
= \(\frac{a \tan x-b}{a \tan x+b}\)
= \(\frac{a\left(\frac{a}{b}\right)-b}{a\left(\frac{a}{b}\right)+b}\)
= \(\frac{a^2-b^2}{a^2+b^2}\)

Question 16.
Is the equation sec2 x – sec x + 1 = 0 possible?
Solution:
Given eqn. be.
6 sec2 x – 5 sec x + 1 = 0
⇒ sec x = \(\)
= \(\)
which is not possible since |sec x| ≥ 1 ∀ x
Hence given eqn. (1) is not possible.

Question 17.
Show that √3 (tan 17° – tan 140°) = 1 + tan 170° tan 140°.
Solution:
Now \(\frac{\tan 170^{\circ}-\tan 140^{\circ}}{1+\tan 170^{\circ} \tan 140^{\circ}}\) = tan (170° – 140°)
[∵ tan (A – B) = \(\frac{\tan A-\tan B}{1+\tan A \tan B}\)]
= tan 30° = \(\frac{1}{\sqrt{3}}\)
⇒ √3 (tan 17° – tan 140°) = 1 + tan 170° tan 140°

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 18.
If A + B = 45°, prove that (1 + tan A) (1 + tan B) = 2. Hence, find the value of tan 22 \(\frac{1}{2}^{\circ}\).
Solution:
Given A + B = 45° ………………(1)
L.H.S. = (1 + tan A) (1 + tan B)
= (1 + tan A) (1 + tan (45° – A)) [using (1)]
= (1 + tan A) \(\left[1+\frac{\tan 45^{\circ}-\tan \mathrm{A}}{1+\tan 45^{\circ} \tan \mathrm{A}}\right]\)
= (1 + tan A) \(\left[1+\frac{1-\tan A}{1+\tan A}\right]\)
= (1 + tan A) \(\left[\frac{1+\tan A+1-\tan A}{1+\tan A}\right]\)
= 2 = R.H.S.

Question 19.
If tan y = \(\frac{Q \sin x}{P+Q \cos x}\), prove that tan (x – y) = \(\frac{P \sin x}{Q+P \cos x}\).
Solution:
Given tan y = \(\frac{Q \sin x}{P+Q \cos x}\) ………………..(1)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 5

Question 20.
Prove that cos2 (x – \(\frac{2 \pi}{3}\)) + cos2 (x + \(\frac{2 \pi}{3}\)) = \(\frac{3}{2}\)
Solution:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 6

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 21.
If x, y and z are in A.P., prove that y = \(\frac{\sin x-\sin z}{\cos z-\cos x}\).
Solution:
Since x, y, z are in A.P.
∴ y – x = z – y
⇒ 2y = x + z

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 7

Question 22.
If sin 2x = λ sin 2y, prove that \(\frac{\tan (x+y)}{\tan (x-y)}=\frac{\lambda+1}{\lambda-1}\).
Solution:
Given sin 2x = λ sin 2y

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 8

Question 23.
Prove that \(\frac{1-\sin 2 x}{1+\sin 2 x}=\tan ^2\left(\frac{\pi}{4}-x\right)\).
Solution:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 9

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 24.
Prove that \(\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8}=\frac{3}{2}\).
Solution:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 10

Question 25.
Given that sin x = – \(\frac{3}{5}\), cos y = – \(\frac{5}{13}\) and x is in the same quadrant asy, evaluate without using tables :
(i) cos (x – y)
(ii) tan (x + y)
(iii) cos \(\frac{x}{2}\)
Solution:
Given sin x = – \(\frac{3}{5}\)
and cos y = – \(\frac{5}{13}\)
Since sin x and cos y both are negative.
∴x and y lies in 3rd quadrant as x and y lies in same quadrant.
So \(\frac{x}{2}\), \(\frac{y}{2}\) lies in second qu1drant.
∴ cos x = – \(\sqrt{1-\sin ^2 x}\)
= \(-\sqrt{1-\left(-\frac{3}{5}\right)^2}\)
= \(-\sqrt{1-\frac{9}{25}}\)
= \(-\sqrt{\frac{16}{25}}=-\frac{4}{5}\)
and sin y = – \(\sqrt{1-\cos ^2 y}\)
= \(-\sqrt{1-\left(\frac{5}{13}\right)^2}\)
= \(-\sqrt{1-\frac{25}{169}}\)
= \(-\sqrt{\frac{144}{169}}=-\frac{12}{13}\)

(i) cos (x – y) = cos x cos y + sin x sin y
= \(\left(-\frac{4}{5}\right)\left(-\frac{5}{13}\right)+\left(-\frac{3}{5}\right)\left(-\frac{12}{13}\right)\)
= \(\frac{20+36}{65}=\frac{56}{65}\)

(ii) tan (x + y) = \(\frac{\tan x+\tan y}{1-\tan x \tan y}\)
= \(\frac{\frac{3}{4}+\frac{12}{5}}{1-\frac{3}{4} \times \frac{12}{5}}\)
= \(\frac{\frac{15+48}{20}}{1-\frac{36}{20}}\)
= \(\frac{63}{-16}\)
[tan x = \(=\frac{\sin x}{\cos x}\)
= \(\frac{-\frac{3}{5}}{-\frac{4}{5}}=\frac{3}{4}\)
and tan y = \(\frac{\sin y}{\cos y}\)
= \(\frac{-\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{5}\)]

(iii) since \(\frac{x}{2}\) lies in second quadrant
∴ cos x < 0
Thus, cos \(\frac{x}{2}\) = – \(\sqrt{\frac{1+\cos x}{2}}\)
= – \(\sqrt{\frac{1+\left(-\frac{4}{5}\right)}{2}}\)
= \(-\sqrt{\frac{1}{10}}=-\frac{1}{\sqrt{10}}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 26.
Prove that :
(i) \(\frac{1-\cos x+\cos y-\cos (x+y)}{1+\cos x-\cos y-\cos (x+y)}=\tan \frac{x}{2} \cot \frac{y}{2}\)
(ii) \(\frac{\cos ^3 x-\cos 3 x}{\cos x}+\frac{\sin ^3 x+\sin 3 x}{\sin x}\) = 3.
Solution:
(i)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 11

(ii)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 12

Question 27.
Prove that \(\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}\) = 4.
Solution:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 13

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test

Question 28.
Show that never lies between \(\frac{1}{3}\) and 3.
Solution:
Let y = \(\frac{\tan 3 x}{\tan x}\)
= \(\frac{3 \tan x-\tan ^3 x}{\tan x\left(1-3 \tan ^2 x\right)}\)
= \(\frac{3-\tan ^2 x}{1-3 \tan ^2 x}\)
⇒ y (1 – 3 tan2 x) = 3 – tan2 x
⇒ tan2 x (1 – 3y) = 3 – y
⇒ tan2 x = \(\frac{3-y}{1-3 y}\)
But tan2 x ≥ 0 ∀ x ∈ R
⇒ \(\frac{3-y}{1-3 y}\) ≥ 0
⇒ (3 – y) (1 – 3y) ≥ 0
[∵ (1 – 3y)2 ≥ 0 ∀ x ∈ R]
⇒ (y – 3) (3y – 1) ≥ 0
⇒ (y – 3) (y – \(\frac{1}{3}\)) ≥ 0
⇒ y ≥ 3 or y ≤ \(\frac{1}{3}\)
[∵ (x – a) (x – b) ≥ 0
and a > b = x ≥ a or x ≤ b]
Thus y i.e. \(\frac{\tan 3 x}{\tan x}\) never lies between \(\frac{1}{3}\) and 3.

Question 29.
Solve the following equations :
(i) tan 2x = – cot (x + \(\frac{\pi}{6}\))
(ii) cot2 x + 3 cosec x + 3 = 0
(iii) 4 sin2 x + √3 = 2 (1 + √3) sin x
(iv) tan2 x – (1 + √3) tan x + √3 = 0
(v) cos 2x – cos 8x + cos 6x = 1
(vi) tan (\(\frac{\pi}{4}\) + x) + tan (\(\frac{\pi}{4}\) – x) = 4
(vii) cosec x = 1 + cot x
Solution:
(i) tan 2x = – cot (x + \(\frac{\pi}{6}\))
⇒ tan 2x = tan \(\left(\frac{\pi}{2}+x+\frac{\pi}{6}\right)\)
= tan \(\left(\frac{2 \pi}{3}+x\right)\)
⇒ 2x = nπ + \(\frac{2 \pi}{3}\) + x ; n ∈ I
[∵ tan θ = tan α
⇒ θ = nπ + α, n ∈ I]
⇒ x = nπ + \(\frac{2 \pi}{3}\) ; n ∈ I

(ii) Given eqn. be,
cot2 x + 3 cosec x + 3 = 0
⇒ cosec2 x – 1 + 3 cosec x + 3 = 0
⇒ cosec2 x + 3 cosec x + 2 = 0
⇒ cosec x = \(\frac{-3 \pm \sqrt{9-8}}{2}\)
= \(\frac{-3 \pm 1}{2}\)
= – 2, – 1
either cosec x = – 2 or cosec x = – 1
sin x = – \(\frac{1}{2}\)
or sin x = – 1
⇒ sin x = – sin \(\frac{\pi}{6}\)
or sin x = sin (- \(\frac{\pi}{2}\))
⇒ x = nπ + (- 1)n (- \(\frac{\pi}{6}\))
or x = nπ + (- 1)n (- \(\frac{\pi}{2}\))
Hence required solutions of given eqn. are
x = nπ – (- 1)n (- \(\frac{\pi}{6}\)) or nπ – (- 1)n (- \(\frac{\pi}{2}\))

(iii) Given eqn. be,
4 sin2 x + √3 = 2 (1 + √3) sin x

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 14

(iv) Given eqn. be,
tan2 x – (1 + √3) tan x + √3 = 0
which is quadratic in tan x.
∴ tan x = \(\frac{(1+\sqrt{3}) \pm \sqrt{(1+\sqrt{3})^2-4 \sqrt{3}}}{2}\)
= \(\frac{(1+\sqrt{3}) \pm \sqrt{(1-\sqrt{3})^2}}{2}\)
= \(\frac{(1+\sqrt{3}) \pm(1-\sqrt{3})}{2}\)
= 1, √3
either tan x = 1 or tan x = √3
tan x = 1 = tan \(\frac{\pi}{4}\)
tan x = √3 = tan \(\frac{\pi}{3}\)
⇒ x = nπ + \(\frac{\pi}{4}\)
or x = nπ + \(\frac{\pi}{3}\) ∀ n ∈ I
Hence required solutions are nπ + \(\frac{\pi}{4}\), nπ + \(\frac{\pi}{3}\) ; n ∈ I

(v) Given eqn.be
cos 2x – cos 8x + cos 6x = 1
⇒cos 2x – cos 8x – (1 – cos 6x) = 0
⇒ 2 sin \(\left(\frac{2 x+8 x}{2}\right)\) sin \(\left(\frac{8 x-2 x}{2}\right)\) – 2 sin2 3x = 0
⇒ 2 sin 5x sin 3x – 2 sin2 3x = 0
⇒ 2 sin 3x [sin 5x – sin 3x] = 0
⇒ 2 sin 3x (2 cos 4x sin x) = 0
either sin 3x = 0 or cos 4x =0 or sin x = 0
3x = nπ or 4x = (2n + 1) \(\frac{\pi}{2}\)
or x = nπ ; n ∈ I
⇒ x = \(\frac{n \pi}{3}\) or x = (2n + 1) \(\frac{\pi}{8}\)
or x = nπ ; n ∈ I
Since the solution x = nπ
i.e. x = π, 2π, 3π ……………….. are included in the
solution x = \(\frac{n \pi}{3}\) for n = 3, 6, 9 ……………..
Hence, the required solutions are
x = nπ, (2n + 1) \(\frac{\pi}{8}\) ∀ n ∈ I

(vi) Given tan \(\left(\frac{\pi}{4}+x\right)\) + tan \(\left(\frac{\pi}{4}-x\right)\) = 4
⇒ \(\frac{\tan \frac{\pi}{4}+\tan x}{1-\tan \frac{\pi}{4} \tan x}+\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \tan x}\) = 4
⇒ \(\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}\) = 4
⇒ (1 + tan x)2 + (1 – tan x)2 = 4 (1 – tan2 x)
⇒ 2 tan2 x + 2 = 4 (1 – tan2 x)
⇒ 6 tan2 x = 2
⇒ tan2 x = \(\frac{1}{3}=\left(\frac{1}{\sqrt{3}}\right)^2\)
= tan2 \(\frac{\pi}{6}\)
⇒ x = nπ ± \(\frac{\pi}{6}\) ∀ n ∈ I
[∵ tan2 θ = tan2 α
⇒ nπ ± α ∀ n ∈ I]

(vii) Given cosec x = 1 + cot x
⇒ \(\frac{1}{\sin x}=1+\frac{\cos x}{\sin x}\)
⇒ 1 = sin x + cos x ; sin x ≠ 0 …………….(1)
which is of the form a cos x + b sin x = c
dividing throughout eqn. (1) by \(\sqrt{1^2+1^2}\) i.e. √2 ; we get
\(\frac{1}{\sqrt{2}}\) cos x + \(\frac{1}{\sqrt{2}}\) sin x = \(\frac{1}{\sqrt{2}}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 15

⇒ x = 2nπ + \(\frac{\pi}{2}\), 2nπ ; n ∈ I
As sin (2n∈) = 0 ∀ n ∈ I but sin x ≠ 0
Hence the required solutions are 2nπ + \(\frac{\pi}{2}\) ; n ∈ I

Question 30.
In any triangle ABC, prove that
(i) 2 (b cos2 \(\frac{C}{2}\) + c cos2 \(\frac{B}{2}\)) = a + b + c
(ii) \(\frac{\sin \mathrm{A}}{\sin (A+B)}=\frac{a}{c}\)
(iii) \(\frac{a-b}{a+b}=\frac{\tan \frac{A-B}{2}}{\tan \frac{A+B}{2}}\)
(iv) \(\frac{b+c}{b-c}=\cot \frac{A}{2} \cot \frac{B-C}{2}\)
(v) \(\frac{1+\cos (\mathrm{A}-\mathrm{B}) \cos \mathrm{C}}{1+\cos (\mathrm{A}-\mathrm{C}) \cos \mathrm{B}}=\frac{a^2+b^2}{a^2+c^2}\)
Solution:
(i) L.H.S. = 2 \(\left(b \cos ^2 \frac{\mathrm{C}}{2}+c \cos ^2 \frac{\mathrm{B}}{2}\right)\)
= 2 \(\left[b\left(\frac{1+\cos \mathrm{C}}{2}\right)+c\left(\frac{1+\cos \mathrm{B}}{2}\right)\right]\)
= [b + c + b cos C + c cos B]
= b + c + a [using projection formulae]
= R.H.S.

(ii) L.H.S. = \(\frac{\sin A}{\sin (A+B)}\)
= \(\frac{\sin A}{\sin (\pi-C)}\)
[∵ A + B + C = π
⇒ A + B = π – C]
= \(\frac{\sin \mathrm{A}}{\sin \mathrm{C}}=\frac{a}{c}\)
[by sine formula, \(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\)
⇒ \(\frac{\sin \mathrm{A}}{\sin \mathrm{C}}=\frac{a}{c}\)]
= R.H.S.

(iii)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 16

(iv)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 19

(v)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 17

Question 31.
In a triangle ABC, if cos A + cos C = sin B, then prove that it is right anglcd triangle.
Solution:
Given cos A + cos C = sin B

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Chapter Test 18

either sin \(\frac{B}{2}\) = 0 or cos C = 0 or cos A = 0
i.e. \(\frac{B}{2}\) = 0 or C = 90° or A = 90°
i.e. B = 0 or C = 90° or A = 90°
either C = 90° or A = 90° [∵ B ≠ 0]
Hence ∆ is right angled triangle.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Students can cross-reference their work with ML Aggarwal Class 11 Solutions Chapter 3 Trigonometry MCQs to ensure accuracy.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Choose the correct answer from the given four options in questions (1 to 44) :

Question 1.
1 radian is approximately equal to
(a) 57°16’
(b) 4718’30”
(c) 53°17’45”
(d) 43°16’
Solution:
(a) 57°16’

We know that,
π radians = 180°

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 1

Question 2.
1° is approximately equal to
(a) 0.001746 radians
(b) 0.01746 radians
(c) 0.0001746 radians
(d) 0.1746 radians
Solution:
(b) 0.01746 radians

We know that,
π radians = 180°
⇒ 1° = \(\frac{\pi}{180}\)
= \(\frac{22}{7 \times 180}\)
⇒ 1° = 0.0 174603 radians.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 3.
If θ lies in second quadrant, then the quadrant in which lies is
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant
Solution:
(d) IV quadrant

Given θ lies in 2nd quadrant
∴ 90° ≤ θ ≤ 180°
⇒ – 45° ≥ – \(\frac{\theta}{2}\) ≥ 90°
⇒ – 90° ≤ – \(\frac{\theta}{2}\) ≤ 45°
∴ θ lies in IV quadrant.

Question 4.
The angle in degree measure between two hands of a clock at 8 : 30 p.m. is
(a) 55°
(b) 66°
(c) 75°
(d) 80°
Solution:
(c) 75°

angle trace out by hour hand in 12 hours = 360°
angle trace out by hour hand in 8\(\frac{1}{2}\) hours = \(\frac{360^{\circ}}{12} \times \frac{17}{2}\)
Φ = 255°
Also, angle trace out by minute hand in 60 minutes = 300°
∴ angle trace out by minute hand in 30 minutes = θ
= \(\frac{360^{\circ}}{60}\) × 30 = 180°
Thus required angle = Φ – θ
= 255° – 180° = 75°

Question 5.
The angle subtended by an arc of length 20 cm at the centre of circle when radius is 14 cm is
(a) \(\frac{5}{7}\) radians
(b) \(\frac{10}{7}\) radians
(c) \(\frac{5}{14}\) radians
(d) \(\frac{7}{10}\) radians
Solution:
(b) \(\frac{10}{7}\) radians

arc length = l = 20 cm ;
r = 14 cm
Since, θ = \(\frac{l}{r}\)
= \(\frac{20}{14}=\frac{10}{7}\) radians

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 6.
A wheel makes 450 revolutions per hour.
The number of radians through which it turns in one second is
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{6}\)
(d) \(\frac{\pi}{3}\)
Solution:
(a) \(\frac{\pi}{4}\)

Given revolutions made by wheel per hour = 450
∴ No. of revolutions made by wheel in one second = \(\frac{450}{60 \times 60}=\frac{1}{8}\)
Further, distance covered by wheel in one revolution = 2π radians
∴ distance covered by wheel in revolution = \(\frac{\pi}{4}\) radians

Question 7.
If sin x = \(\frac{3}{5}\), then cosx is
(a) \(\frac{4}{5}\) but not \(-\frac{4}{5}\)
(b) \(\frac{4}{5}\) or \(-\frac{4}{5}\)
(c) \(-\frac{4}{5}\) but not \(\frac{4}{5}\)
(d) none of these
Solution:
(b) \(\frac{4}{5}\) or \(-\frac{4}{5}\)

Given sin x = (b) \(\frac{3}{5}\)
∴ cos x = ± \(\sqrt{1-\sin ^2 x}\)
= ± \(\sqrt{1-\frac{9}{25}}= \pm \frac{4}{5}\)

Question 8.
The value of cosec (- 750°) is
(a) \(\frac{2}{\sqrt{3}}\)
(b) – 2
(c) 2
(d) – \(\frac{2}{\sqrt{3}}\)
Solution:
(c) 2

cosec (- 750°) = – cosec 750°
[∵ cosec (- θ) = – cosec θ]
= – cosec (720° + 30°)
= – cosec 30°
= – 2

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 9.
The value of tan (- \(\frac{15 \pi}{4}\)) is
(a) – 1
(b) 1
(c) \(\frac{1}{\sqrt{3}}\)
(d) \(-\frac{1}{\sqrt{3}}\)
Solution:
(b) 1

tan (- \(\frac{15 \pi}{4}\)) = – tan \(\frac{15 \pi}{4}\)
[∵ tan (- θ) = – tan θ]
= – tan (2π + \(\frac{7 \pi}{4}\))
= – tan \(\frac{7 \pi}{4}\)
= – tan (2π – \(\frac{\pi}{4}\))
= – {- tan \(\frac{\pi}{4}\)}
= tan \(\frac{\pi}{4}\) = 1

Question 10.
The range of 4 + 5 cos x is
(a) [- 1, 9]
(b) (- 1, 9]
(c) (- 1, 9)
(d) [- 1, 9)
Solution:
(a) [- 1, 9]

Since – 1 ≤ cos x ≤ 1
⇒ – 5 ≤ 5 cos x ≤ 5
⇒ 4 – 5 ≤ 4 + 5 cos x ≤ 4 + 5
⇒ – 1 ≤ f (x) ≤ 9
∴ Rf = [- 1, 9]

Question 11.
The domain of 2 sin x cos x is
(a) R – πn
(b) R – (2n + 1) \(\frac{\pi}{2}\)
(c) R
(d) R – \(\frac{n \pi}{2}\)
Solution:
(c) R

Let f (x) = 2 sin x cos x = sin 2x, which is defined ∀ x ∈ R
∴ Df = R

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 12.
If x lies in III quadrant and tan x = \(\frac{5}{12}\), then sin x and cos x respectively are
(a) \(\frac{5}{13}, \frac{12}{13}\)
(b) \(-\frac{5}{13}, \frac{12}{13}\)
(c) \(\frac{5}{13},-\frac{12}{13}\)
(d) \(-\frac{5}{13},-\frac{12}{13}\)
Solution:
(d) \(-\frac{5}{13},-\frac{12}{13}\)

Given tan x = \(\frac{5}{12}\)
Since x lies in IIIrd quadrant
∴ sin x, cos x < 0
∴ sec x = – \(\sqrt{1+\tan ^2 x}\)
= – \(\sqrt{1+\left(\frac{5}{12}\right)^2}\)
= \(-\sqrt{1+\frac{25}{144}}=-\frac{13}{12}\)
cos x = – \(\frac{12}{13}\)
∴ sin x = tan x cos x
= \(\frac{5}{12} \times\left(-\frac{12}{13}\right)\)
= – \(\frac{5}{13}\)

Question 13.
The value of 2 sin2 \(\frac{\pi}{6}\) + cosec \(\frac{7 \pi}{6}\) . cos2 \(\frac{\pi}{32}\) is equal to
(a) 1
(b) \(\frac{3}{2}\)
(c) – 1
(d) 2
Solution:
(b) \(\frac{3}{2}\)

2 sin2 \(\frac{\pi}{6}\) + cosec \(\frac{7 \pi}{6}\) . cos2 \(\frac{\pi}{32}\) = \(2\left(\sin \frac{\pi}{6}\right)^2+\left\{\ {cosec}\left(\pi+\frac{\pi}{6}\right)\right\}^2\left(\cos \frac{\pi}{3}\right)^2\)
= \(2\left(\frac{1}{2}\right)^2+\left\{-\ {cosec} \frac{\pi}{6}\right\}^2\left(\frac{1}{2}\right)^2\)
= \(\frac{1}{2}+(-2)^2 \times \frac{1}{4}=\frac{3}{2}\)

Question 14.
The value of \(\cos \left(\frac{3 \pi}{2}+x\right) \cdot \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]\) is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

cos (\(\frac{3 \pi}{2}\) + x) . cos (2π + x) [cot (\(\frac{3 \pi}{2}\) – x) + cot (2π + x)]
= sin x . cos x [tan x + cot x]
= sin x cos x \(\left[\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right]\)
= \(\frac{\sin x \cos x\left(\sin ^2 x+\cos ^2 x\right)}{\sin x \cos x}\) = 1

Question 15.
If sin θ + cosec θ = 2, then sin2 θ + cosec2 θ is equal to
(a) 1
(b) 4
(c) 2
(d) 6
Solution:
(c) 2

Given sin θ + cosec θ = 2 …………………….( 1)
∴ sin2 θ + cosec2 θ = (sin θ + cosec θ)2 – 2 sin θ cosec θ
= 22 – 2 = 2 [using eqn. (1)]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 16.
If f (x) = cos2 x + sec2 x, then
(a) f (x)< 1
(b) f(x) = 1
(c) 2< f (x) < 1
(d) f (x) ≥ 2
Solution:
(d) f (x) ≥ 2

f(x) = cos2 x + sec2 x
= (cos x – sec x)2 + 2 cos x sec x
= (cos x – sec x)2 + 2 ≥ 2
[∵ (cos x – sec x)2 ≥ 0 ∀ x ∈ R]

Question 17.
If tan θ = \(\frac{1}{2}\) and tan = \(\frac{1}{3}\), then the value of (θ + Φ) is
(a) \(\frac{\pi}{6}\)
(b) π
(c) 0
(d) \(\frac{\pi}{4}\)
Solution:
(d) \(\frac{\pi}{4}\)

tan (θ + Φ) = \(\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}\)
= \(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\)
= \(\frac{\frac{5}{6}}{1-\frac{1}{6}}\)
= \(\frac{\frac{5}{6}}{\frac{5}{6}}\)
= 1
θ + Φ = \(\frac{\pi}{4}\)

Question 18.
Which of the following is not correct?
(a) sin θ = – \(\frac{1}{5}\)
(b) cos θ = 1
(c) sec θ = \(\frac{1}{2}\)
(d) tan θ = 20
Solution:
(c) sec θ = \(\frac{1}{2}\)

since |sec θ| ≥ 1
∴ sec θ = \(\frac{1}{2}\) is not possible.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 19.
The value of cos 1°. cos 2°. cos 3°………………. cos 179° is
(a) \(\frac{1}{\sqrt{2}}\)
(b) 0
(c) 1
(d) – 1
Solution:
(b) 0

cos 1°. cos 2°………………. cos 90°………………. cos 179° = 0
[∵ cos 90° = 0]

Question 20.
The value of tan 10. tan 2°. tan 3° ……………… tan 89° is
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) 2
Solution:
(b) 1

(tan 1° tan 2° tan 3° ………………. tan 44°) tan 45° (tan 46° tan 47° …………… tan 89°)
= (tan 1° tan 2° …. tan 44°) × 1 × (tan (90° – 44°) tan (90° – 43°) …. tan (90° – 1°)]
= (tan 1° tan 2° ………………. tan 44°) (cot 44° cot 43° ……………….. cot 1°)
= (tan 1° cot 1°) (tan 2° cot 2°) ……………. (tan 44° cot 44°) –
= 1 × 1 × ……………. × 1 = 1
[∵ tan θ . cot θ = 1]

Question 21.
The value of tan 75° – cot 75° is equal to
(a) 2√3
(b) 2 + √3
(c) 2 – √3
(d) 1
Solution:

tan 75° – cot 75° = tan 75°
= tan 75° – \(\frac{1}{\tan 75^{\circ}}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 2

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 22.
Which of the following is correct?
(a) sin 1° > sin 1
(b) sin 1° < sin 1
(c) sin 1° = sin 1
(d) sin 1° = \(\frac{\pi}{180}\) sin 1
Solution:
(b) sin 1° < sin 1

Since 1 radian = 57° 16’
Thus 1° < 1 radian
⇒ sin 1° < sin 1
[∵ since sin θ be an increasing function in first quadrant]

Question 23.
The value of tan 5A – tan 3A – tan 2A is equal to
(a) tan 5A . tan 3A. tan 2A
(b) – tan 5A . tan3A . tan 2A
(c) tan 3A. tan 2A – tan 2A. tan 5A – tan 5A tan 2A
(d) none of these
Solution:
(a) tan 5A . tan 3A. tan 2A

tan 5A = tan (3A + 2A)
= \(\frac{\tan 3 \mathrm{~A}+\tan 2 \mathrm{~A}}{1-\tan 3 \mathrm{~A} \cdot \tan 2 \mathrm{~A}}\)
= tan 5A (1 – tan 3A tan 2A)
= tan 3A + tan 2A
= tan 5A – tan 5A – tan 2A
= tan 5A tan 3A tan 2A

Question 24.
The value of sin (45° + θ) – cos (45° – θ) is
(a) 2 cos θ
(b) 2 sin θ
(c) 1
(d) 0
Solution:
(d) 0

sin (45° + θ) – cos (45° – θ) = sin 45° cos θ + cos 45° sin θ – [cos 45° cos θ + sin 45° sin θ]
[∵ sin (A + B) = sin A cos B + cos A sin B
and cos (A – B) = cos A cos B + sin A sin B]
= \(\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta\)
= 0

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 25.
The value of cot (\(\frac{\pi}{4}\) + θ) . cot (\(\frac{\pi}{4}\) – θ) is
(a) – 1
(b) 0
(c) 1
(d) not defined
Solution:
(c) 1

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 3

Question 26.
cos 2θ . cos 2Φ + sin2 (θ – Φ) – sin2 (θ + Φ) is equal to
(a) sin 2 (θ + Φ)
(b) cos 2 (θ + Φ)
(c) sin 2 (θ – Φ)
(d) cos 2 (- Φ)
Solution:
(b) cos 2 (θ + Φ)

cos 2θ cos 2Φ + sin2 (θ – Φ) – sin2 (θ + Φ) = cos 2θ cos 2Φ + sin (θ – Φ + θ + Φ) sin (θ – Φ – θ – Φ)
[∵ sin2 A – sin2 B = sin (A + B) sin (A – B)]
= cos 2θ cos 2Φ + sin 2θ sin (- 2Φ)
= cos 2θ cos 2Φ – sin 2θ sin 2Φ
= cos (2θ + 2Φ)

Question 27.
If for real x, cos θ = x + \(\frac{1}{x}\), then
(a) θ is an acute angle
(b) is right angle
(c) θ is an obtuse angle
(d) no value of θ is possible
Solution:
(d) no value of θ is possible

Given cos θ = x + \(\frac{1}{x}\)
⇒ x2 – x cos θ + 1 = 0
Here D = cos2 θ – 4 < 0
[∵ 0 ≤ cos2 θ ≤ 1]
∴ x is non-real, which contradicts the given fact that x is real.
Thus there is no value of θ for which the given equation holds.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 28.
If tan A = \(\frac{1}{2}\), tan B = \(\frac{1}{3}\) then tan (2A + B) is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(c) 3

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 4

Question 29.
If α + β = \(\frac{\pi}{3}\), then (1 + tan α) (1 + tan β) is
(a) 1
(b) 2
(c) – 1
(d) – 2
Solution:
(b) 2

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 5

Question 30.
If sin θ = – \(\frac{4}{5}\) and θ lies in third quadrant, then the value of cos \(\frac{\theta}{2}\) is
(a) \(\frac{1}{5}\)
(b) \(-\frac{1}{\sqrt{10}}\)
(c) \(-\frac{1}{\sqrt{5}}\)
(d) \(\frac{1}{\sqrt{10}}\)
Solution:
(c) \(-\frac{1}{\sqrt{5}}\)

Since 180° ≤ θ ≤ 270°
⇒ 9o° ≤ \(\frac{\theta}{2}\) ≤ 135°
∴ \(\frac{\theta}{2}\) lies in 2nd quadrant
∴ cos \(\frac{\theta}{2}\) < 0
given sin θ = – \(\frac{4}{5}\) and θ lies in third quadrant
∴ cos θ < 0

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 6

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 31.
The greatest value of sin x . cos x ¡s
(a) 1
(b) 2
(c) √2
(d) \(\frac{1}{2}\)
Solution:
(d) \(\frac{1}{2}\)

sin x cos x = \(\frac{1}{2}\) (2 sin x cos x)
= \(\frac{1}{2}\) sin 2x
= f (x)
since – 1 ≤ sin 2x ≤ 1
⇒ – \(\frac{1}{2}\) ≤ \(\frac{1}{2}\) sin 2x ≤ \(\frac{1}{2}\)
– \(\frac{1}{2}\) ≤ f (x) ≤ \(\frac{1}{2}\)
∴ greatest value of f (x) = \(\frac{1}{2}\)

Question 32.
a cos x + b sin x lies between
(a) a and b
(b) – (a2 + b2) and (a2 + b2)
(c) \(-\sqrt{a^2+b^2} \text { and } \sqrt{a^2+b^2}\)
(d) \(-\sqrt{a+b} \text { and } \sqrt{a+b}\)
Solution:
(c) \(-\sqrt{a^2+b^2} \text { and } \sqrt{a^2+b^2}\)

put a = r cos α
and b = r sin α
On squaring and adding eqn. (1) and (2) ; we have
a2 + b2 = r2
⇒ r = \(\sqrt{a^2+b^2}\) (∵ r > 0)
On dividing eqn. (2) by eqn. (1) ; we have
tan α = \(\frac{b}{a}\)
∴ a cos x + b sin x = r cos α cos x + r sin x sin α
= r cos (x – α)
since – 1 ≤ cos (x – α) ≤ 1
⇒ – r ≤ r cos (x – α) ≤ r
⇒ – \(\sqrt{a^2+b^2}\) ≤ a cos x + b sin x ≤ \(\sqrt{a^2+b^2}\).

Question 33.
Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2π] is
(a) 0
(b) 1
(c) 2
(d) 3
Solution
(c) 2

Given eqn. be, tan x + sec x 2 cos x
⇒ \(\frac{\sin x+1}{\cos x}\) = 2 cos x
⇒ sin x + 1 = 2 cos2 x, cosx ≠ 0
⇒ 1 + sin x = 2 (1 – sin2 x), x ≠ \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
2 sin2 x + sin x – 1 = 0
⇒ sin x = \(\frac{-1 \pm \sqrt{1+8}}{4}\)
= \(\frac{-1 \pm 3}{4}\)
⇒ sin x = \(\frac{1}{2}\), – 1 ; x ≠ \(\frac{\pi}{2}\), \(\frac{3 \pi}{2}\)
∴ sin x = \(\frac{1}{2}\) = \(\frac{\pi}{6}\), sin (π – \(\frac{\pi}{6}\))
[∵ x ∈ [0, 2π])
and sin x = – 1
= sin \(\frac{3 \pi}{2}\), sin (- \(\frac{\pi}{2}\)) but x ∈ [0, 2π]
⇒ x = \(\frac{3 \pi}{2}\), – \(\frac{3 \pi}{2}\) which are not possible.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 34.
If cos x = – \(\frac{3}{5}\) and π < x < \(\frac{3 \pi}{2}\), then \(\) is equal to (a) \(\frac{1}{6}\) (b) – \(\frac{1}{3}\) (c) – \(\frac{1}{6}\) (d) \(\frac{2}{3}\) Solution: Given cos x = – \(\frac{3}{5}\) and x lies in 3rd quadrant. ∴ tan x, cot x > 0 and cosec x < 0
∴ sec x = \(\frac{1}{\cos x}=-\frac{5}{3}\)
and tan x = \(\sqrt{\sec ^2 x-1}\)
= \(\sqrt{\frac{25}{9}-1}=\frac{4}{3}\)
and cot x = \(\frac{3}{4}\) ;
sin x = tan x . cos x
= \(\frac{4}{3} \times\left(-\frac{3}{5}\right)=-\frac{4}{5}\)
∴ cosec x = – \(\frac{5}{4}\)
Thus, \(\frac{\ {cosec} x+\cot x}{\sec x-\tan x}=\frac{-\frac{5}{4}+\frac{3}{4}}{-\frac{5}{3}-\frac{4}{3}}\)
= \(\frac{-\frac{2}{4}}{-\frac{9}{3}}=\frac{1}{6}\)

Question 35.
The value of sin (π + x). sin (π – x) . coscc2 x is
(a) 0
(b) 1
(c) – \(\frac{1}{6}\)
(d) \(\frac{2}{3}\)
Solution:
(c) – \(\frac{1}{6}\)

sin (π + x) sin(π – x) . cosec2 x = – sin x sin x cosec2 x = – 1

Question 36.
The value of \(3 \sin \frac{\pi}{6} \cdot \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cdot \cot \frac{\pi}{4}\) is
(a) – 1
(b) 0
(c) 1
(d) 2
Solution:
(c) 1

\(3 \sin \frac{\pi}{6} \cdot \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cdot \cot \frac{\pi}{4}\)
= \(3 \times \frac{1}{2} \times \frac{2}{1}-4 \sin \left(\pi-\frac{\pi}{6}\right) \times 1\)
= 3 – 4 sin \(\frac{\pi}{6}\)
= 3 – 4 sin × \(\frac{1}{2}\)
= 3 – 2 = 1

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 37.
If sin α = k sin β, then tan \(\left(\frac{\alpha-\beta}{2}\right)\) . cot \(\left(\frac{\alpha+\beta}{2}\right)\) is equal to
(a) \(\frac{k-1}{k+1}\)
(b) \(\frac{k+1}{k-1}\)
(c) \(\frac{1+k}{1-k}\)
(d) \(\frac{1-k}{1+k}\)
Solution:
(a) \(\frac{k-1}{k+1}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 7

Question 38.
If sin x = \(\frac{1}{3}\), then the value of sin 3x is
(a) 1
(b) 0
(c) \(\frac{23}{27}\)
(d) – \(\frac{23}{27}\)
Solution:

Given sin x = \(\frac{1}{3}\)
Thus, sin 3x = 3 sin x – 4 sin3 x
= \(3 \times \frac{1}{3}-4\left(\frac{1}{3}\right)^3\)
= \(1-\frac{4}{27}=\frac{23}{27}\)

Question 39.
The value of cos2 48° – sin2 12° is
(a) \(\frac{\sqrt{5}+1}{8}\)
(b) \(\frac{\sqrt{5}-1}{8}\)
(c) \(\frac{\sqrt{5}+1}{4}\)
(d) \(\frac{\sqrt{5}-1}{4}\)
Solution:

cos2 48° – sin2 12° = cos (48° + 12°) cos (48° – 12°)
[∵ cos (A + B) cos (A – B) = cos2 A – sin2 B]
= cos 60° cos 36°
= \(\frac{1}{2} \times \frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}+1}{8}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 40.
The value of \(\cos \frac{2 \pi}{15} \cdot \cos \frac{4 \pi}{15} \cdot \cos \frac{8 \pi}{15} \cdot \cos \frac{16 \pi}{15}\) is
(a) \(-\frac{1}{16}\)
(b) \(-\frac{1}{8}\)
(c) \(\frac{1}{8}\)
(d) \(\frac{1}{16}\)
Solution:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 8

Question 41.
The general solution of the equation tan (2x + \(\frac{\pi}{12}\)) = 0 is
(a) \(\frac{n \pi}{2}-\frac{\pi}{24}\)
(b) nπ + \(\frac{\pi}{12}\)
(c) nπ – \(\frac{\pi}{12}\)
(d) \(\frac{n \pi}{2}+\frac{\pi}{24}\)
Solution:
(a) \(\frac{n \pi}{2}-\frac{\pi}{24}\)

Given tan(2x + \(\frac{\pi}{12}\)) = 0
⇒ 2x + \(\frac{\pi}{12}\) = nπ
[∵ tan θ = 0
⇒ θ = nπ]
⇒ 2x = nπ – \(\frac{\pi}{12}\)
⇒ x = \(\)

Question 42.
The general solution of the equation sin2 x . sec x + √3 tan x = 0 is
(a) (2n + 1) \(\frac{\pi}{2}\)
(b) nπ
(c) nπ + \(\frac{\pi}{6}\)
(d) nπ – \(\frac{\pi}{6}\)
Solution:

Given eqn. be,
sin2 x sec x + √3 tan x = 0
⇒ \(\frac{\sin ^2 x}{\cos x}+\frac{\sqrt{3} \sin x}{\cos x}\) = 0
⇒ sin2 x + √3 sin x = 0, cos x ≠ 0
⇒ sin x (sin x + √3) = 0, cos x ≠ 0
⇒ sin x = 0 or sin x = – √3, cos x ≠ 0
but |sin x| ≤ 1
∴ sin x = – √3 is not possible.
∴ sin x = 0
⇒ x = nπ and x ≠ nπ + \(\frac{\pi}{2}\)
⇒ x = nπ, n ∈ I

Question 43.
If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of 2 cot A – 5 cos A + sin A is
(a) \(-\frac{53}{10}\)
(b) \(\frac{23}{10}\)
(c) \(\frac{37}{10}\)
(d) \(\frac{7}{10}\)
Solution:
3 tan A + 4 = 0
⇒ tan A = – \(\frac{4}{3}\)
Since A lies in 2nd quadrant
∴ sin A > 0, cos A < 0
sec A = – \(\sqrt{1+\tan ^2 A}\)
= – \(\sqrt{1+\frac{16}{9}}=-\frac{5}{3}\)
⇒ cos A = – \(\frac{3}{5}\)
and sin A = tan A . cos A
= \(-\frac{4}{3} \times\left(-\frac{3}{5}\right)=\frac{4}{5}\)
∴ 2 cot A – 5 cos A + sin A = \(2\left(-\frac{3}{4}\right)-5\left(-\frac{3}{5}\right)+\frac{4}{5}\)
= \(-\frac{3}{2}+3+\frac{4}{5}\)
= \(\frac{-15+30+8}{10}=\frac{23}{10}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs

Question 44.
If tan θ = \(\frac{a}{b}\), then b cos 2θ + a sin 2θ is equal to
(a) a
(b) b
(c) \(\frac{a}{b}\)
(d) \(\frac{b}{a}\)
Solution:
(b) b

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry MCQs 9

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Access to comprehensive Class 11 ISC Maths Solutions Chapter 3 Trigonometry Ex 3.9 encourages independent learning.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Very short answer/objective questions (1 to 4) :

Question 1.
In a ∆ABC, if a = 4, b = 5 and c = 6, find cos C.
Solution:
Given a = 4, b = 5 and c = 6 using cosine formula, we have
cos C = \(\frac{a^2+b^2-c^2}{2 a b}\)
= \(\frac{4^2+5^2-6^2}{2 \times 4 \times 5}\)
= \(\frac{16+25-36}{40}\)
= \(\frac{5}{40}=\frac{1}{8}\)

Question 2.
In a ∆ABC, if a = 7, c = 8 and ∠B = 45°, find area of ∆ABC.
Solution:
Given a = 7 ; c = 8 ; ∠B = 45°
∴ area of ∆ABC = \(\frac{1}{2}\) × a × c
= \(\frac{1}{2}\) × 7 × 8 × sin 45°
= \(\frac{28}{\sqrt{2}}\)
= 14√2 sq. units

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 1

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Question 3.
In a ∆ABC, if a = 4, b = 6 and c = 8, find the value of 4 cos B + 3 cos C.
Solution:
Given a = 4, b = 6, c = 8
∴ cos B = \(\frac{a^2+c^2-b^2}{2 a c}\)
= \(\frac{4^2+8^2-6^2}{2 \times 4 \times 8}\)
= \(\frac{80-36}{64}\)
= \(\frac{44}{64}=\frac{11}{16}\)
and cos C = \(\frac{a^2+b^2-c^2}{2 a b}\)
= \(\frac{4^2+6^2-8^2}{2 \times 4 \times 6}\)
= \(-\frac{12}{48}=-\frac{1}{4}\)
Thus 4 cos B + 3 cos C = \(4 \times \frac{11}{16}+3\left(-\frac{1}{4}\right)\)
= \(\frac{11-3}{4}\)
= 2

Question 4.
In a ∆ABC, if a = 3, b = 5 and c = 6, then verify that c = a cos B + b cos A.
Solution:
Given a = 3, b = 5 and c = 6
R.H.S = a cos B + b cos A
= \(a\left(\frac{a^2+c^2-b^2}{2 a c}\right)+b\left(\frac{b^2+c^2-a^2}{2 b c}\right)\)
= \(\frac{1}{2 c}\) [a2 + c2 – b2 + b2 + c2 – a2]
= \(\frac{2 c^2}{2 c}\)
= c = L.H.S

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Short and long answer questions (5 to 25) :

Question 5.
2(b ccos A + ca cos B + ab cos C) = a2 + b2 + c2.
Solution:
LH.S. = 2 (bc cos A + ca cos B + ah cos C)
= 2 \(\left[b c\left(\frac{b^2+c^2-a^2}{2 b c}\right)+c a\left(\frac{c^2+a^2-b^2}{2 c a}\right)+a b\left(\frac{a^2+b^2-c^2}{2 a b}\right)\right]\)
[using cosine’s formula]
= \(\frac{2}{2}\) [b2 + c2 – a2 – c2 + a2 – b2 + a2 + b2 – c2]
= a2 + b2 + c2
= R.H.S.

Question 6.
\(\frac{\cos \mathrm{A}}{a}+\frac{\cos \mathrm{B}}{b}+\frac{\cos \mathrm{C}}{c}=\frac{a^2+b^2+c^2}{2 a b c}\)
Solution:
L.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 2

Question 7.
(a + b) sin \(\frac{C}{2}\) = c cos \(\frac{A-B}{2}\)
Solution:
In any ∆ABC, using sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k (say)
⇒ a = k sin A ;
b = k sin B
and c = k sin C

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 3

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Question 8.
(a – b) cos \(\frac{C}{2}\) = c sin \(\frac{A-B}{2}\)
Solution:
In any ∆ABC, using sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k (say)
⇒ a = k sin A ;
b = k sin B
and c = k sin C

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 4

Question 9.
(c – b)2 cos2 \(\frac{A}{2}\) + (c + b)2 sin2 \(\frac{A}{2}\) = a2.
Solution:
L.H.S. = (c – b)2 cos2 \(\frac{A}{2}\) + (c + b)2 sin2 \(\frac{A}{2}\)
= (b2 + c2) \(\left(\cos ^2 \frac{\mathrm{A}}{2}+\sin ^2 \frac{\mathrm{A}}{2}\right)\) – 2 bc \(\left(\cos ^2 \frac{\mathrm{A}}{2}-\sin ^2 \frac{\mathrm{A}}{2}\right)\)
= (b2 + c2) × 1 – 2bc cos A
= (b2 + c2) – 2bc \(\left(\frac{b^2+c^2-a^2}{2 b c}\right)\)
= b2 + c2 – b2 – c2 + a2
= a2
= R.H.S.

Question 10.
\(\frac{a^2 \sin (B-C)}{\sin B+\sin C}+\frac{b^2 \sin (C-A)}{\sin C+\sin A}+\frac{c^2 \sin (A-B)}{\sin A+\sin B}\) = 0
Solution:
L.H.S. = \(\frac{a^2 \sin (B-C)}{\sin B+\sin C}+\frac{b^2 \sin (C-A)}{\sin C+\sin A}+\frac{c^2 \sin (A-B)}{\sin A+\sin B}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 5

= k2 sin A (sin B – sin C) + k2 sin B (sin C – sin A) + k2 sin C (sin A – sin B)
= k2 [sin A sin B – sin A sin C + sin B sin C – sin A sin B + sin C sin A – sin B sin C]
= k2 × 0
= 0 = R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Question 11.
\(\frac{a^2-b^2}{\cos A+\cos B}+\frac{b^2-c^2}{\cos B+\cos C}+\frac{c^2-a^2}{\cos C+\cos A}\) = 0.
Solution:
L.H.S. = \(\frac{a^2-b^2}{\cos A+\cos B}+\frac{b^2-c^2}{\cos B+\cos C}+\frac{c^2-a^2}{\cos C+\cos A}\)
In any ∆ ABC, bby sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k (say)
⇒ a = k sin A ;
b = k sin B
and c = k sin C
∴ L.H.S. = \(\frac{k^2\left[\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}\right]}{\cos \mathrm{A}+\cos \mathrm{B}}+\frac{k^2\left[\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}\right]}{\cos \mathrm{B}+\cos \mathrm{C}}+\frac{k^2\left[\sin ^2 \mathrm{C}-\sin ^2 \mathrm{~A}\right]}{\cos \mathrm{C}+\cos \mathrm{A}}\)
= \(\frac{k^2\left[\cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A}\right]}{\cos \mathrm{B}+\cos \mathrm{A}}+\frac{k^2\left[\cos ^2 \mathrm{C}-\cos ^2 \mathrm{~B}\right]}{\cos \mathrm{C}+\cos \mathrm{B}}+\frac{k^2\left[\cos ^2 \mathrm{~A}-\cos ^2 \mathrm{C}\right]}{\cos \mathrm{A}+\cos \mathrm{C}}\)
[∵ sin2 θ + cos2 θ = 1
sin2 θ = 1 – cos2 θ]
= k2 [cos B – cos A + cos C – cos B + cos A – cos C]
= k2 × 0
= 0 = R.H.S.

Question 12.
b2 sin 2C + c2 sin 2B = 2bc sin A
Solution:
L.H.S. = b2 sin 2C + c2 sin 2B
= 2b2 sin C cos C + 2c2 sin B cos B
By sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k
⇒ a = k sin A ;
b = k sin B
and c = k sin C
∴ L.H.S. = 2b2 × \(\frac{c}{k}\left(\frac{b^2+a^2-c^2}{2 a b}\right)\) + 2c2 × \(\frac{b}{k} \frac{\left(c^2+a^2-b^2\right)}{2 a c}\)
= \(\frac{b c}{a k}\) (b2 + a2 – c2) + \(\frac{b c}{a k}\) (c2 + a2 – b2)
= \(\frac{b c}{a k}\) [b2 + a2 – c2 + c2 + a2 – b2]
= \(\frac{b c}{a k}\) × 2a2
= \(\frac{2 a b c}{k}\)
= 2bc \(\left(\frac{a}{k}\right)\)
= 2bc sin A [using sine formula]
= R.H.S.

Question 13.
\(\frac{b^2-c^2}{a} \cos \mathrm{A}+\frac{c^2-a^2}{b} \cos \mathrm{B}+\frac{a^2-b^2}{c} \cos \mathrm{C}\) = 0
Solution:
L.H.S. = \(\frac{b^2-c^2}{a} \cos \mathrm{A}+\frac{c^2-a^2}{b} \cos \mathrm{B}+\frac{a^2-b^2}{c} \cos \mathrm{C}\)
= \(\frac{b^2-c^2}{a}\left(\frac{b^2+c^2-a^2}{2 b c}\right)+\frac{c^2-a^2}{b}\left(\frac{c^2+a^2-b^2}{2 a c}\right)+\frac{a^2-b^2}{c}\left(\frac{a^2+b^2-c^2}{2 a b}\right)\)
[using cosine’s formula]
= \(\frac{1}{2 a b c}\) [(b2 – c2) (b2 + c2 – a2) + (c2 – a2) (c2 + a2 – b2) + (a2 – b2) (a2 + b2 – c2)]
= \(\frac{1}{2 a b c}\) [b4 – c4 – a2 (b2 – c2) + c4 – a4 – b2 (c2 – a2) + a4 – b4 – c2 (a2 – b2)]
= \(\frac{1}{2 a b c}\) [- a2b2 + a2c2 – b2c2 + a2b2 – c2a2 + b2c2]
= \(\frac{1}{2 a b c}\) × 0
= 0 = R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Question 14.
b2 cos 2A – a2 cos 2B = b2 – a2.
Solution:
Using sine formula
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k
⇒ a = k sin A ;
b = k sin B
and c = k sin C
L.H.S. = b2 cos 2A – a2 cos 2B
= b2 (1 – 2 sin2 A) – a2 (1 – 2 sin2 B)
= b2 – a2 – 2b2 \(\left(\frac{a}{k}\right)^2\) + 2a2 \(\left(\frac{b}{k}\right)^2\)
= b2 – a2 – \(\frac{2 b^2 a^2}{k^2}+\frac{2 a^2 b^2}{k^2}\)
= b2 – a2
= R.H.S.

Question 15.
(c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C.
Solution:
Using sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k
⇒ a = k sin A ;
b = k sin B
and c = k sin C
⇒ a = k sin A ;
b = k sin B
and c = k sin C
Now (c2 – a2 + b2) tan A

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 6

Thus,
(c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C
[using (1), (2) and (3)].

Question 16.
2 [a sin2 \(\frac{C}{2}\) + c sin2 \(\frac{A}{2}\)] = c + a – b
Solution:
LH.S. = \(\left[a \sin ^2 \frac{\mathrm{C}}{2}+c \sin ^2 \frac{\mathrm{A}}{2}\right]\)
= 2 \(\left[a\left(\frac{1-\cos \mathrm{C}}{2}\right)+c\left(\frac{1-\cos \mathrm{A}}{2}\right)\right]\)
= a + c – a cos C – c cos A
= a + c – b [using projection formula)
= R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Question 17.
If in a triangle ABC, sin 2A + sin 2B = sin 2C, prove that either A = 90° or B = 90°.
Solution:
Given sin 2A + sin 2B = sin 2C
⇒ 2 sin (A + B) cos (A – B) = 2 sin C cos C
⇒ sin [π – C] cos (A – B) = sin C cos C
[∵ A + B + C = π
⇒ A + B = π – C]
⇒ sin C [cos (A – B) – cos C] = 0
⇒ sin C [cos (A – B) – cos (π – \(\overline{A+B}\))] = 0
⇒ sin C [cos(A – B) cos (A + B)] = 0
⇒ sin C [2 cos A cos B] = 0
either sin C = 0 or cos A = 0 or cos B = 0
C = 0 or A = 90° or B = 90°
since A, B and C are the angles of ∆ABC
∴ neither of the angles A, B and C is 0.
Hence either A = 90° or B = 90°.

Question 18.
If in a triangle ABC, sin2 A + sin2 B = sin2 C. prove that ¿ABC is right angled.
Solution:
Given sin2 A + sin2 B = sin2 C ………………..(1)
using sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k
⇒ a = k sin A ;
b = k sin B
and c = k sin C
Thus from (1) ; we have
\(\frac{a^2}{k^2}+\frac{b^2}{k^2}=\frac{c^2}{k^2}\)
⇒ a2 + b2 = c2
So the triangles ABC is right angled ∆ at C.
since pythagoras theorem holds.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 7

Question 19.
If in a triangle ABC, \(\frac{\cos \mathrm{A}}{a}=\frac{\cos \mathrm{B}}{b}\), show that the triangle is isosceles.
Solution:
Given, \(\frac{\cos \mathrm{A}}{a}=\frac{\cos \mathrm{B}}{b}\)
⇒ b cos A = a cos B ………………..(1)
In any ABC, using sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k
∴ from (1) ;
k sin B cos A = k sin A cos B
⇒ 2 sin B cosA = 2 sin A cos B
⇒ sin (A + B) – sin (A – B) = sin (A + B) + sin (A – B)
⇒ 2 sin (A – B) = 0
⇒ A – B = 0
⇒ A = B
Hence the given triangle is isosceless.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Question 20.
In triangle ABC, if a = 18, b = 24 and c = 30 then find
(i) cos A, cos B, cos C
(ii) sin A sin B, sin C.
Solution:
(i) Given a = 18, b = 24 ; c = 30
Using cosine formula, we have

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 8

(ii) Using sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\)
⇒ \(\frac{18}{\sin A}=\frac{24}{\sin B}=\frac{30}{\sin \frac{\pi}{2}}\)
⇒ sin A = \(=\frac{18}{30}=\frac{3}{5}\) ;
sin B = \(\frac{24}{30}=\frac{4}{5}\)
and sin C = 1

Question 21.
If a = 7 cm, b = 5 cm and c = 3 cm, prove that the triangle ha 14ij angle.
Solution:
Given a = 7 cm ;
b = 5 cm
and c = 3 cm
∴ cos A = \(\frac{b^2+c^2-a^2}{2 b c}\)
= \(\frac{5^2+3^2-7^2}{2 \times 5 \times 3}\)
= \(\frac{34-49}{30}\)
= \(-\frac{15}{30}=-\frac{1}{2}\)
cos A = – cos \(\frac{\pi}{3}\)
= cos (π – \(\frac{\pi}{3}\))
= cos \(\frac{2 \pi}{3}\)
⇒ A = \(\frac{2 \pi}{3}\)
Clearly one of the angle of given ∆ be 120° > 90°.
Hence one of the angle ofi be obtuse angle.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Question 22.
If the angles of a triangle are in the ratio 1 : 2 : 3, prove that the corresponding sides of the triangle are in ratio 1 : √3 : 2.
Solution:
Let the angles of triangle be x°, 2x° and 3x°
since, x° + 2x° + 3x° = 180°
⇒ 6x° = 180°
⇒ x = 330°
Hence the angles of a triangle are 30°, 60° and 90°.
Using sine formula, we have
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\)
⇒ \(\frac{a}{\sin 30^{\circ}}=\frac{b}{\sin 60^{\circ}}=\frac{c}{\sin 90^{\circ}}\)
⇒ \(\frac{a}{\frac{1}{2}}=\frac{b}{\frac{\sqrt{3}}{2}}=\frac{c}{1}\)
⇒ \(\frac{a}{1}=\frac{\dot{b}}{\sqrt{3}}=\frac{c}{2}\)
Hence a : b : c :: 1 : √3 : 2
Thus, the corresponding sides of the triangle are in the ratio 1 : √3 : 2.

Question 23.
In a ABC, a = 1, b = √3 and C = \(\frac{\pi}{6}\). Find the other two angles and the third side.
Solution:
Given a = 1 ;
b = √3
and C = \(\frac{\pi}{6}\) or 30°
In only ∆ABC, using sine formula, we have

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 9

⇒ 3 = 4 – c2
⇒ c2 = 1
⇒ c = 1 [∵ c > 0]
∴ from (1) ;
\(\frac{1}{\sin A}=\frac{\sqrt{3}}{\sin B}=\frac{1}{\frac{1}{2}}\)
⇒ sin A = \(\frac{1}{2}\)
⇒ A = \(\frac{\pi}{6}\)
Since A. B and C are the angles of ∆ABC.
∴ A + B + C = 180°
⇒ B = 180° – 30° – 30° = 120°

Question 24.
Two boats leave a place at the same time. One travels 56 km in the direction N 40° E, while the other travels 48 km in the direction S 80° E. What is the distance between the boats?
Solution:
Clearly ∠POQ = 180° – 40° – 80° = 60°
using cosine’s formula, we have
cos 60° = \(\frac{\mathrm{OP}^2+\mathrm{OQ}^2-\mathrm{PQ}^2}{2 \mathrm{OP} \cdot \mathrm{OQ}}\)
⇒ \(\frac{1}{2}=\frac{56^2+48^2-\mathrm{PQ}^2}{2 \times 56 \times 48}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 10

⇒ 56 × 48 = 562 + 482 – PQ2
⇒ PQ2 = 3136 + 2304 – 2688 = 2752
⇒ PQ = \(\sqrt{2752}\) cm = 52.5 km

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Question 25.
Two trees A and B are on the same side of the river. From a point C in the river the distance of the trees A and B is 250 m and 300 m respectively. 1f the angle C is 45, find the distance between the trees. Use √2 = 1.44.
Solution:
Using cosine formula, we have
cos C = \(\frac{a^2+b^2-c^2}{2 a b}\)
= \(\frac{\mathrm{CA}^2+\mathrm{CB}^2-\mathrm{AB}^2}{2 \mathrm{CA} \cdot \mathrm{CB}}\)
⇒ cos 45° = \(\frac{(250)^2+(300)^2-\mathrm{AB}^2}{2 \times 250 \times 300}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9 11

⇒ \(\frac{1}{\sqrt{2}}\) × 2 × 250 × 300 = 62500 + 90000 – AB2
⇒ AB2 = 152500 – 75000 √2
⇒ AB2 = 152500 – 106050 = 46450
[∵ √2 = 1.414]
⇒ AB = \(\sqrt{46450}\) = 215.5 m
Hence, the required distance between trees be 215.5 metre.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Students appreciate clear and concise ISC Maths Class 11 Solutions Chapter 3 Trigonometry Ex 3.8 that guide them through exercises.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 1.
Find the principal solutions of the following equations :
(i) sin x = \(\frac{\sqrt{3}}{2}\)
(ii) cos x = 1
(iii) tan x = \(\frac{1}{\sqrt{3}}\)
(iv) tan x = – \(\frac{1}{\sqrt{3}}\)
(v) cos x = – 1
(vi) cosec x = – 1
(vii) sec x = – √2
(viii) tan x = – 1
(ix) cosec x = – 2.
Solution:
(i) We know that, the solutions lying between 0 to 2π (0 ≤ x < 2π) are principal solutions.
Given sin x = \(\frac{\sqrt{3}}{2}\)
we know that,
sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
and sin (π – \(\frac{\pi}{2}\)) = sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
∴ sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
and sin \(\frac{2 \pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
thus the principal solutions are \(\frac{\pi}{3}\), \(\frac{2 \pi}{3}\).

(ii) Given cos x = 1
since cos 0 = 1
and cos 2π = 1
but 2π ∉ [0, 2π)
∴ The principal soln. be 0.

(iii) Given tan x = \(\frac{1}{\sqrt{3}}\)
We know that,
tan \(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)
and tan (π + \(\frac{\pi}{6}\)) = tan \(\frac{\pi}{6}\)
= \(\frac{1}{\sqrt{3}}\)
∴ tan \(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)
and tan (\(\frac{7 \pi}{6}\)) = \(\frac{1}{\sqrt{3}}\)
Thus, the principal solutions are \(\frac{\pi}{6}\) and \(\frac{7 \pi}{6}\).

(iv) Given tan x = – \(\frac{1}{\sqrt{3}}\)
We know that,
tan \(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)
∴ tan (π – \(\frac{\pi}{6}\)) = – tan \(\frac{\pi}{6}\)
= – \(\frac{1}{\sqrt{3}}\)
and tan (2π – \(\frac{\pi}{6}\)) = – tan \(\frac{\pi}{6}\)
= – \(\frac{1}{\sqrt{3}}\)
Thus tan \(\frac{5 \pi}{6}\) = – \(\frac{1}{\sqrt{3}}\)
and tan \(\left(\frac{11 \pi}{6}\right)=-\frac{1}{\sqrt{3}}\)
Hence, \(\frac{5 \pi}{6}\) and \(\frac{11 \pi}{6}\) are the two principal solutions.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

(v) Given cos x = – 1
We know that,
cos 0 = 1
∴ cos (π – 0) = – cos 0 = – 1
Thus, the only principal soln. be π.

(vi) Given cosec x = – 1
⇒ sin x = – 1
We know that,
sin \(\frac{\pi}{2}\) = 1
∴ sin (π + \(\frac{\pi}{2}\)) = – sin \(\frac{\pi}{2}\) = – 1
or sin (2π – \(\frac{\pi}{2}\)) = – sin \(\frac{\pi}{2}\) = – 1
Thus, sin \(\frac{3 \pi}{2}\) = – 1
Hence \(\frac{3 \pi}{2}\) be the only principal soln.

(vii) Given sec x = – √2
⇒ cos x = – \(\frac{1}{\sqrt{2}}\)
We know that,
cos \(\frac{\pi}{4}=\frac{1}{\sqrt{2}}\)
∴ cos (π – \(\frac{\pi}{4}\)) = – cos \(\frac{\pi}{4}\)
= – \(\frac{1}{\sqrt{2}}\)
and cos (π + \(\frac{\pi}{4}\)) = – cos \(\frac{\pi}{4}\)
= – \(\frac{1}{\sqrt{2}}\)
Thus, cos \(\frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}\)
and cos \(\frac{5 \pi}{4}=-\frac{1}{\sqrt{2}}\)
Hence, \(\frac{3 \pi}{4}\) and \(\frac{5 \pi}{4}\) are the principal solutions.

(viii) Given tan x = – 1 …( 1)
we know that,
tan \(\frac{\pi}{4}\) = 1
tan (π – \(\frac{\pi}{4}\)) = – tan \(\frac{\pi}{4}\) = – 1
and tan (2π – \(\frac{\pi}{4}\)) = – tan \(\frac{\pi}{4}\) = – 1
Thus, tan \(\frac{3 \pi}{4}\) = – 1
and tan (\(\frac{7 \pi}{4}\)) = – 1
Hence the principal solutions of eqn. (1) are \(\frac{3 \pi}{4}\) and \(\frac{7 \pi}{4}\).

(ix) Given √3 cosec x = – 2
⇒ cosec x = – \(\frac{2}{\sqrt{3}}\)
We know that
sin x = – \(\frac{\sqrt{3}}{2}\)
We know that,
sin \(\frac{\pi}{3}\) = \(\frac{\sqrt{3}}{2}\)
sin (π + \(\frac{\pi}{3}\)) = – sin \(\frac{\pi}{3}\)
= – \(\frac{\sqrt{3}}{2}\)
and sin (2π – \(\frac{\pi}{3}\)) = – sin \(\frac{\pi}{3}\)
= – \(\frac{\sqrt{3}}{2}\)
Thus sin \(\frac{4 \pi}{3}\) = – \(\frac{\sqrt{3}}{2}\)
and sin \(\frac{5 \pi}{3}=-\frac{\sqrt{3}}{2}\)
Hence the required solution of given eqn. are \(\frac{4 \pi}{3}\) and \(\frac{5 \pi}{3}\).

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Find the general solutions of the following (2 to 18) equations :

Question 2.
(i) sin 3x = 0
(ii) sin \(\frac{5 x}{2}\) = 0
(iii) cos (x + \(\frac{\pi}{12}\)) = 0
(iv) cos (x + \(\frac{\pi}{4}\)) = 0
(v) tan \(\frac{3 x}{4}\) = 0
(vi) tau (x – \(\frac{x}{4}\)) = 0
Solution:
(i) Given sin 3x = 0 = sin 0
⇒ 3x = n π ∀ n ∈ I
x = \(\frac{n \pi}{4}\) ∀ n ∈ I

(ii) Given sin \(\frac{5 x}{2}\) = 0
⇒ \(\frac{5 x}{2}\) = n π, n ∈ I
⇒ x = \(\frac{2 n \pi}{5}\) ∀ n ∈ I

(iii) Given sin (x + \(\frac{\pi}{12}\)) = 0
⇒ x + \(\frac{\pi}{12}\) = nπ ∀ n ∈ I
[∵ sin θ = 0
⇒ θ = nπ ∀ n ∈ I]
⇒ x = nπ – \(\frac{\pi}{12}\), n ∈ I

(iv) cos (x + \(\frac{\pi}{4}\)) = 0
x + \(\frac{\pi}{4}\) = (2n + 1) \(\frac{\pi}{2}\), n ∈ I
⇒ x = (2n + 1) \(\frac{\pi}{2}\) – \(\frac{\pi}{4}\), n ∈ I
⇒ x = nπ + \(\frac{\pi}{4}\), n ∈ I

(v) Given tan \(\frac{3x}{4}\) = 0
⇒ \(\frac{3x}{4}\) = nπ, n ∈ I
⇒ x = \(\frac{4 n \pi}{3}\), n ∈ I

(vi) Given tan (x – \(\frac{\pi}{4}\)) = 0
⇒ x – \(\frac{\pi}{4}\) = nπ ∀ n ∈ I
⇒ x = nπ + \(\frac{\pi}{4}\) ∀ n ∈ I

Question 3.
(i) cos x = \(\frac{\sqrt{3}}{2}\)
(ii) cot x = \(\frac{1}{\sqrt{3}}\)
(iii) sec x = 2
(iv) sin x = – \(\frac{\sqrt{3}}{2}\)
(v) cot x = – √3
(vi) cosec x = – 2.
Solution:
(i) Given cos x = \(\frac{\sqrt{3}}{2}\)
⇒ cos x = cos \(\frac{\pi}{6}\)
⇒ x = 2nπ ± \(\frac{\pi}{6}\) ∀ n ∈ I
[∵ cos θ = cos α
⇒ θ = 2nπ ± α, n ∈ I]
which is the required general soln.

(ii) Given cot x = \(\frac{1}{\sqrt{3}}\)
⇒ tan x = √3 = tan \(\frac{\pi}{3}\)
⇒ x = nπ + \(\frac{\pi}{3}\) ∀ n ∈ I
which is the required soln.

(iii) Given sec x = 2
⇒ cos x = \(\frac{1}{2}\)
= cos \(\frac{\pi}{3}\)
⇒ x = 2nπ ± \(\frac{\pi}{3}\) ∀ n ∈ I

(iv) Given sin x = – \(\frac{\sqrt{3}}{2}\)
= – sin \(\frac{\pi}{3}\)
= sin (π + \(\frac{\pi}{3}\))
⇒ sin x = sin \(\frac{4 \pi}{3}\)
⇒ x = nπ + (- 1)n \(\frac{4 \pi}{3}\) ∀ n ∈ I
[∵ sin θ = sin α
⇒ θ = nπ + (- 1)n \(\frac{4 \pi}{3}\) ∀ n ∈ I]

(v) Given cot x = – √3
⇒ tan x = – \(\frac{1}{\sqrt{3}}\)
= – tan \(\frac{\pi}{2}\)
= tan (π – \(\frac{\pi}{6}\))
⇒ tan x = tan \(\frac{5 \pi}{6}\)
⇒ x = nπ + \(\frac{5 \pi}{6}\) ∀ n ∈ I
[∵ tan θ = tan α
⇒ θ = nπ + α ∀ n ∈ I]

(vi) Given cosec x = – 2
⇒ sin x = – \(\frac{1}{2}\)
= – sin \(\frac{\pi}{6}\)
= sin (π + \(\frac{\pi}{6}\))
⇒ sin x = sin \(\frac{7 \pi}{6}\)
⇒ x = nπ + (- 1)n \(\frac{7 \pi}{6}\) ∀ n ∈ I

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 4.
(i) √2 sin x + 1 = 0
(ii) 1 + tan x = 0
(iii) √3 sec x + 2 = 0
Solution:
(i) Given √2 sin x + 1 = 0
⇒ sin x = – \(\frac{1}{\sqrt{2}}\)
= – sin \(\frac{\pi}{4}\)
= sin (π + \(\frac{\pi}{4}\))
⇒ sin x = sin \(\frac{5 \pi}{4}\)
⇒ x = nπ + (- 1)n \(\frac{5 \pi}{4}\) ; n ∈ I
which is the required solution.

(ii) Given 1+ tan x = 0
⇒ tan x = – 1
= – tan \(\frac{\pi}{4}\)
⇒ tan x = – tan (π – \(\frac{\pi}{4}\))
⇒ tan x = tan \(\frac{\pi}{4}\)
⇒ x = nπ + \(\frac{3 \pi}{4}\) where n ∈ I

(iii) Given √3 sec x + 2 = 0
⇒ sec x = – \(\frac{2}{\sqrt{3}}\)
⇒ cos x = – \(\frac{\sqrt{3}}{2}\)
⇒ cos x = – cos \(\frac{\pi}{6}\)
= cos (π – \(\frac{\pi}{6}\))
⇒ cos x = cos \(\frac{5 \pi}{6}\)
⇒ x = 2nπ ± \(\frac{5 \pi}{6}\) ; ∀ n ∈ I

Question 5.
(i) tan2 (x + \(\frac{\pi}{3}\)) = 0
(ii) 4 cos2 x = 3
(iii) 3 cosec2 x = 4.
Solution:
(i) Given tan2 (x + \(\frac{\pi}{3}\)) = 0
= tan2 0
⇒ x + \(\frac{\pi}{3}\) = nπ – \(\frac{\pi}{3}\) ∀ n ∈ I
[∵ tan2 θ = tan2 α
⇒ θ = nπ ± α ; n ∈ I]
⇒ x = nπ – \(\frac{\pi}{3}\) ∀ n ∈ I

(ii) Given 4 cos2 x = 3
⇒ cos2 x = \(\frac{3}{4}\)
= \(\left(\frac{\sqrt{3}}{2}\right)^2\)
= cos2 \(\frac{\pi}{6}\)
⇒ x = nπ ± \(\frac{\pi}{6}\) ; n ∈ I
[∵ cos2 θ = cos2 α
⇒ θ = nπ ± α ; n ∈ I]

(iii) Given 3 cosec2 x = 4
cosec2 x = \(\frac{4}{3}\)
sin2 x = \(\frac{3}{4}=\left(\frac{\sqrt{3}}{2}\right)^2\)
sin2 x = sin2 \(\frac{\pi}{3}\)
x = nπ ± \(\frac{\pi}{3}\) ; n ∈ I

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 6.
(i) sin 2x = \(\frac{1}{2}\)
(ii) cos 3x= – \(\frac{1}{2}\)
(iii) tan \(\frac{2 x}{3}\) = √3.
Solution:
(i) Given sin 2x = \(\frac{1}{2}\)
= sin \(\frac{\pi}{6}\)
⇒ 2x = nπ + (- 1)n \(\frac{\pi}{6}\)
⇒ x = \(\frac{n \pi}{2}\) + (- 1)n \(\frac{\pi}{12}\) ∀ n ∈ I

(ii) Given cos 3x = – \(\frac{1}{2}\)
=- cos \(\frac{\pi}{3}\)
= cos (π – \(\frac{\pi}{3}\))
⇒ cos 3x = cos \(\frac{2 \pi}{3}\)
⇒ 3x = 2nπ ± \(\frac{2 \pi}{3}\)
⇒ x = \(\frac{2 n \pi}{3} \pm \frac{2 \pi}{9}\) ∀ n ∈ I

(iii) Given tan \(\frac{2 x}{3}\) = √3
⇒ \(\tan \frac{2 x}{3}=\tan \frac{\pi}{3}\)
⇒ \(\frac{2 x}{3}\) = nπ + \(\frac{\pi}{3}\) ∀ n ∈ I
⇒ x = \(\frac{3 n \pi}{2}+\frac{\pi}{2}\) ; ∀ n ∈ I

Question 7.
(i) sec 3x = – √2
(ii) cot 4x = – 1
(iii) cosec 3x = – \(\frac{2}{\sqrt{3}}\)
Solution:
(i) Given sec 3x = – √2
⇒ cos 3x = \(-\frac{1}{\sqrt{2}}=-\cos \frac{\pi}{4}\)
= cos (π – \(\frac{\pi}{4}\))
⇒ cos 3x = cos \(\frac{3 \pi}{4}\)
⇒ 3x = 2nπ ± \(\frac{3 \pi}{4}\) ∀ n ∈ I
⇒ x = \(\frac{2 n \pi}{3} \pm \frac{\pi}{4}\) ; n ∈ I

(ii) Given cot 4x = – 1
⇒ tan 4x = – 1
= – tan \(\frac{\pi}{4}\)
= tan (π – \(\frac{\pi}{4}\))
⇒ tan 4x = tan \(\frac{3 \pi}{4}\)
⇒ 4x = nπ + \(\frac{3 \pi}{4}\)
⇒ x = \(\frac{n \pi}{4}+\frac{3 \pi}{16}\) ∀ n ∈ I

(iii) cosec 3x = – \(\frac{2}{\sqrt{3}}\)
⇒ sin 3x = – \(\frac{\sqrt{3}}{2}\)
= – sin \(\frac{\pi}{3}\)
= sin (π + \(\frac{\pi}{3}\))
⇒ sin 3x = sin \(\frac{4 \pi}{3}\)
⇒ 3x = nπ + (- 1)n \(\frac{4 \pi}{3}\) ∀ n ∈ I
⇒ x = \(\frac{n \pi}{3}+(-1)^n \frac{4 \pi}{9}\) ∀ n ∈ I

Question 8.
(i) cos 3x = cos x
(ii) sin 9x sin x.
Solution:
(i) Given, cos 3x = cos x
⇒3x = 2nπ ± x ∀ n ∈ I
⇒ 3x = 2nπ + x ∀ n ∈ I
or 3x = 2nπ – x ∀ n ∈ I
⇒ x = nπ or x = \(\frac{n \pi}{2}\) ∀ n ∈ I
Also, we note that, the solutions
X = nπ ∀ n ∈ I
i.e. x = π, 2π, 3π are included in the solutions
∴ x = \(\frac{n \pi}{2}\) when n = 2, 4, 6, ……………….
Thus, the required solutions are x = \(\frac{n \pi}{2}\) ∀ n ∈ I

(ii) Given, sin 9x = sin x
⇒ 9x = nπ + (- 1)n x ∀ n ∈ I

Case – I.
When n be even integer
i.e. n = 2m
9x = 2mπ + x
⇒ x = \(\frac{m \pi}{4}\) ∀ n ∈ I

Case – II :
When n be on odd integer
i.e. n = 2m + 1
9x = (2m + 1) π – x
⇒ x = (2m + 1) \(\frac{\pi}{10}\) ∀ n ∈ I
Hence, x = \(\frac{m \pi}{4}\), (2m + 1) \(\frac{\pi}{10}\) ∀ n ∈ I

Aliter :
Given sin 9x – sin x = 0
⇒ 2 cos \(\left(\frac{9 x+x}{2}\right)\) sin \(\left(\frac{9 x-x}{2}\right)\) = 0
⇒ 2 cos 5x sin 4x = 0
either cos 5x = 0 or sin 4x = 0
⇒ 5x = (2n + 1) \(\frac{\pi}{2}\) or 4x = nπ
⇒ x = (2n + 1) \(\frac{\pi}{10}\)
or x = \(\frac{n \pi}{4}\) ∀ n ∈ I
Hence the required solutions are (2n + 1)\(\frac{\pi}{10}\) and nπ, where n ∈ I

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 9.
(i) tan 3x = cot x
(ii) tan x + cot 3x = 0
Solution:
(i) Given tan 3x = cot x
= tan (\(\frac{\pi}{2}\) – x)
⇒ 3x = nπ + \(\frac{\pi}{2}\) – x
⇒ 4x = nπ + \(\frac{\pi}{2}\)
⇒ x = \(\frac{n \pi}{4}+\frac{\pi}{8}\), where n ∈ I
which ¡s the required general soln.

(ii) Given tan x + cot 3x = 0
⇒ tan x = – cot 3x
= tan (\(\frac{\pi}{2}\) + 3x)
⇒ x = nπ + \(\frac{\pi}{2}\) + 3x
[∵ tan θ = tan α
⇒ θ = nπ + α]
⇒ – 2x = nπ + \(\frac{\pi}{2}\)
⇒ x = \(-\frac{n \pi}{2}-\frac{\pi}{4}\)
⇒ x = \(\frac{m \pi}{2}-\frac{\pi}{4}\) ; where m ∈ I

Question 10.
(i) 2 sin x cos x = sin x
(ii) tan2 x – √3 tan x = 0.
Solution:
(i) Given 2 sin x cos x = sin x
⇒ sin x (2 cos x – 1) = 0
either sin x = 0 or 2 cos x – 1 = 0
⇒ x = nπ or x = \(\frac{1}{2}\), n ∈ I
⇒ x = nπ or cos x = cos \(\frac{\pi}{3}\), n ∈ I
⇒ x = nπ or x = 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I
Hence, the required solutions are nπ, 2nπ ± \(\frac{\pi}{3}\); n ∈ I

(ii) tan2 x – √3 tan x = 0
⇒ tan x (tan x – √3) = 0
either tan x = 0 or tan x = √3
x = nπ or tan x = tan \(\frac{\pi}{3}\) ; n ∈ I
x = nπ or x = nπ + \(\frac{\pi}{3}\) ∀ n ∈ I
Hence the required solutions are nit,Hence the required solutions are nπ, nπ + \(\frac{\pi}{3}\); where n ∈ I

Question 11.
(i) sin x = tan x
(ii) sec2 x = 1 + tan x.
Solution:
(i) Given sin x = tan x
⇒ sin x = \(\frac{\sin x}{\cos x}\)
⇒ sin x cos x – sin x = 0 [∵ cos x ≠ 0]
⇒ sin x (cos x – 1) = 0
⇒ sin x = 0 or cos x = 1 = cos 0
⇒ x = – nπ or x = 2nπ ± 0 ∀ n ∈ I
⇒ x = nπ or x = 2nπ ∀ n ∈ I
We note that the solutions x = 2nπ, where n ∈ I
i.e. x = 0, 2,π 4π, 6π are included in x = mπ
For n = 0, 2, 4, ……………….
Hence the required solution be nπ where n ∈ I

(ii) Given sec2 x = 1 + tan x
⇒ 1 + tan2 x = 1 + tan x
⇒ tan x (tan x – 1) = 0
either tan x = 0 or tan x = 1
⇒ x = nπ or tan x = tan \(\frac{\pi}{4}\)
⇒ x = nπ or x = nπ + \(\frac{\pi}{4}\) ∀ n ∈ I
Thus nπ, nπ + \(\frac{\pi}{4}\) ; n ∈ I are the required solutions of given equation.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 12.
(i) cosec2 x = 2 cot x
(ii) tan x + cot x = 2.
Solution:
(i) Given, cosec2 x = 2 cot x
⇒ 1 + cot2 x – 2 cot x = 0
⇒ (cot x – 1)2 = 0
⇒ cot x = 1
⇒ tan x = 1
⇒ tan x = tan \(\frac{\pi}{4}\)
⇒ x = nπ + \(\frac{\pi}{4}\) where n ∈ I

(ii) Given tan x + cot x = 2
⇒ tan x + \(\frac{1}{\tan x}\) = 2
⇒ tan2 x – 2 tan x + 1 = 0
[∵ tan x ≠ 0]
⇒ (tan x – 1)2 = 0
⇒ tan x = 1 = tan \(\frac{\pi}{4}\)
⇒ x = nπ + \(\frac{\pi}{4}\) ; where n ∈ I

Question 13.
(i) tan x + cot x = 2 sec x
(ii) cos 3x + cos x – 2 cos 2x = 0.
Solution:
(i) Given tan x + cot x = 2 sec x
⇒ \(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=\frac{2}{\cos x}\)
⇒ \(\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}=\frac{2}{\cos x}\)
⇒ \(\frac{1}{\sin x \cos x}=\frac{2}{\cos x}\)
⇒ 2 sin x = 1
[∵ cos x ≠ 0]
⇒ sin x = \(\frac{1}{2}\)
= sin \(\frac{\pi}{6}\)
⇒ x = nπ + (- 1)n \(\frac{\pi}{6}\) ∀ n ∈ I

(ii) Given cos 3x + cos x – 2 cos 2x = 0
⇒ 2 cos \(\left(\frac{3 x+x}{2}\right)\) cos \(\left(\frac{3 x-x}{2}\right)\) – 2 cos 2x = 0
[∵ cos C + cos D = 2 cos \(\left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\) cos \(\left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)]
⇒ 2cos 2x cos x – 2 cos 2x = 0
⇒ 2 cos 2x (cos x – 1) = 0
either cos 2x = 0 or cos x – 1 = 0
i.e. 2x = (2n +1) \(\frac{\pi}{2}\) or cos x = 1 = cos 0
⇒ x = (2n + 1) \(\frac{\pi}{4}\) or x = 2nπ ; where n ∈ I

Question 14.
(i) sin 2x + sin 4x + sin 6x = 0
(ii) sin x – sin 2x + sin 3x = 0
Solution:
(i) Given, sin 2x + sin 4x + sin 6x = 0
⇒ (sin 6x + sin 2x) + sin 4x = 0
⇒ 2 sin \(\left(\frac{6 x+2 x}{2}\right)\) cos \(\left(\frac{6 x-2 x}{2}\right)\) + sin 4x = 0
⇒ 2 sin 4x cos 2x + sin 4x = 0
⇒ sin 4x (2 cos 2x + 1) = 0
either sin 4x = 0
or cos 2x = – \(\frac{1}{2}\) = – cos \(\frac{\pi}{3}\)
⇒ 4x = nπ or cos 2x = + cos (π – \(\frac{\pi}{3}\))
⇒ 4x = nπ or 2x = 2nπ ± \(\frac{2 \pi}{3}\) ∀ n ∈ I
⇒ x = \(\frac{n \pi}{4}\) or x = nπ ± \(\frac{\pi}{3}\) ; n ∈ I
Thus the required solutions are \(\frac{n \pi}{4}\) or nπ ± \(\frac{\pi}{3}\), n ∈ I

(ii) Given sin x – sin 2x + sin 3x = 0
⇒ (sin 3x + sin x) – sin 2x = 0
⇒ 2 sin \(\left(\frac{3 x+x}{2}\right)\) cos \(\left(\frac{3 x-x}{2}\right)\) – sin 2x = 0
⇒ 2 sin 2x cos x – sin 2x = 0
⇒ sin 2x (2 cos x – 1) = 0
either sin 2x = 0
or cos x = \(\frac{1}{2}\)
= cos \(\frac{\pi}{3}\)
⇒ 2x = nπ or x = 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I
Thus, required solutions are \(\frac{n \pi}{2}\) or 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 15.
(i) 7 cos2 x + 3 sin2 x = 4
(ii) Zsin2x=3cosx.
Solution:
(i) Given eqn. be,
7 cos2 x + 3 sin2 x = 4
⇒ 7 cos2 x + 3 (1 – cos2 x) = 4
⇒ 4 cos2 x = 1
⇒ cos2 x = \(\frac{1}{4}=\left(\frac{1}{2}\right)^2\)
= cos2 \(\frac{\pi}{3}\)
⇒ x = nπ ± \(\frac{\pi}{3}\), where n ∈ I
[∵ cos2 x = cos2 α
⇒ x = nπ ± α where n ∈ I]

(ii) Given, 2 sin2 x = 3 cos x
⇒ 2 (1 cos2 x) = 3 cos x
⇒ 2 cos2 x + 3 cos x – 2 = 0
⇒ cos x = \(\frac{-3 \pm \sqrt{9+16}}{4}\)
= \(\frac{-3 \pm 5}{4}\)
= \(\frac{1}{2}\), – 2
Since cos x = – 2 is not possible
[∵ – 1 ≤ cos x ≤ 1]
Thus, cos x = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
⇒ x = 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I

Question 16.
(i) 4 cos2 x – 4 sin x – 1 = 0
(ii) 2 sin2 x + √3 cos x + 1 = 0
Solution:
(i) Given 4 cos2 x – 4 sin x – 1 = 0
⇒ 4(1 – sin2 x) – 4 sin x – 1 = 0
⇒ 4 sin2 x + 4 sin x – 3 = 0
⇒ sin x = \(\frac{-4 \pm \sqrt{16+48}}{8}\)
= \(\frac{-4 \pm 8}{8}\)
= \(\frac{1}{2},-\frac{3}{2}\)
But sin x = – \(\frac{3}{2}\) is not possible
[∵ |sin x| ≤ 1]
Thus sin x = \(\frac{1}{2}\) = sin \(\frac{\pi}{6}\)
⇒ x = nπ + (- 1)n \(\frac{\pi}{6}\) ; where n ∈ I

(ii) Given, 2 sin2 x + √3 cos x + 1 = 0.
⇒ 2 (1 – cos2 x) + cos x + 1 = 0
⇒ 2 cos2 x – √3 cos x – 3 = 0
⇒ cos x = \(\frac{\sqrt{3} \pm \sqrt{3+24}}{4}\)
= \(\frac{\sqrt{3} \pm 3 \sqrt{3}}{4}\)
cos x = \(\frac{4 \sqrt{3}}{4}, \frac{-2 \sqrt{3}}{4}\)
cos x = √3, \(\frac{-\sqrt{3}}{2}\)
But cos x = √3 is not possible.
[∵ |cos x| ≤ 1]
Thus, cos x = \(\frac{-\sqrt{3}}{2}\)
= – cos \(\frac{\pi}{6}\)
= cos (π – \(\frac{\pi}{6}\))
⇒ cos x = cos \(\frac{5 \pi}{6}\)
⇒ x = 2nπ ± \(\frac{5 \pi}{6}\) ; n ∈ I

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 17.
(i) cos x + sin x = 1
(ii) cos x – sin x+ 1 = 0
(iii) √3 cos x – sin x = 1
Solution:
(i) Given cos x + sin x = 1 ………………(1)
which is of the form a cos x + b sin x = C
Dividing throughout eqn. (1) by \(\sqrt{a^2+b^2}\)
i.e. \(\sqrt{1^2+1^2}\) i.e. √2 ; we have

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8 1

(ii) Given cos x – sin x + 1 = 0
⇒ sin x – cos x = 1
⇒ cos x – sin x = – 1 …………………..(1)
which is of the form a cos x + b sin x = c
So dividing throughout eqn. (1) by
\(\sqrt{1^2+(-1)^2}\) i.e. by √2 ; we have

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8 2

(iii) Given eqn. be,
√3 cos x – sin x = 1
⇒ \(\frac{\sqrt{3}}{2} \cos x-\frac{1}{2} \sin x=\frac{1}{2}\)
⇒ cos x cos \(\frac{\pi}{6}\) – sin x sin \(\frac{\pi}{6}\) = \(\frac{1}{2}\)
⇒ cos (x + \(\frac{\pi}{6}\)) = cos \(\frac{\pi}{3}\)
⇒ x + \(\frac{\pi}{6}\) = 2nπ ± \(\frac{\pi}{3}\)
[∵ cos x = cos α
⇒ x = 2nπ ± α where n ∈ I]
⇒ x = 2nπ ± \(\frac{\pi}{3}-\frac{\pi}{6}\)
⇒ x = 2nπ + \(\frac{\pi}{6}\), 2nπ – \(\frac{\pi}{2}\), where n ∈ I

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.8

Question 18.
(i) √2 sec x + tan x = 1
(ii) cosec x = cot x + √3
Solution:
(i) Given √2 sec x + tan x = 1
⇒ \(\frac{\sqrt{2}}{\cos x}+\frac{\sin x}{\cos x}\) = 1
⇒ √2 + sin x = cos x ; cos x ≠ 0
⇒ cos x – sin x = √2, cos x ≠ 0
which is of the form a cos x + b sin x = c
So dividing throughout eqn. (1) by \(\sqrt{1^2+(-1)^2}\) i.e. by √2 ; we have
\(\frac{1}{\sqrt{2}}\) cos x – \(\frac{1}{\sqrt{2}}\) sin x = 1
⇒ cos \(\frac{\pi}{4}\) cos x – sin \(\frac{\pi}{4}\) sin x = 1
⇒ cos (x + \(\frac{\pi}{4}\)) = cos 0
⇒ x + \(\frac{\pi}{4}\) = 2nπ ∀ n ∈ I
⇒ x = 2nπ – \(\frac{\pi}{4}\) ∀ n ∈ I

(ii) cosec x = cot x + √3
⇒ \(\frac{1}{\sin x}=\frac{\cos x}{\sin x}\) + √3
⇒ 1 = cos x + √3 sin x, sin x ≠ 0 ……………….(1)
which n of the form a cos x + b sin x = c
So dividing throughout eqn. (1) by \(\sqrt{1^2+(\sqrt{3})^2}\) i.e. 2 ; we have
i.e. \(\frac{1}{2}\) cos x + \(\frac{\sqrt{3}}{2}\) sin x = \(\frac{1}{2}\)
⇒ cos \(\frac{\pi}{3}\) cos x + sin \(\frac{\pi}{3}\) sin x = \(\frac{1}{2}\)
⇒ cos (x – \(\frac{\pi}{3}\)) = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
⇒ x – \(\frac{\pi}{3}\) = 2nπ ± \(\frac{\pi}{3}\) ; n ∈ I
⇒ x = 2nπ ± \(\frac{\pi}{3}\) + \(\frac{\pi}{3}\) ∀ n ∈ I
i.e. x = 2nπ + \(\frac{\pi}{3}\), 2nπ where n ∈ I
But sin (2nπ) = 0 so we reject these values.
Hence the required solutions are
x = 2nπ + \(\frac{2 \pi}{3}\) ∀ n ∈ I

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Interactive ISC Maths Class 11 Solutions Chapter 3 Trigonometry Ex 3.7 engage students in active learning and exploration.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Very short answer/objective questions (1 to 3) :

Question 1.
If sin x = \(\frac{2}{3}\), find the value of cos 2x.
Solution:
Given sin x = \(\frac{2}{3}\)
∴ cos 2x = 1 – 2 sin2 x
= 1 – 2 \(\left(\frac{2}{3}\right)^2\)
= \(1-2 \times \frac{4}{9}\)
= \(1-\frac{8}{9}=\frac{1}{9}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 2.
If cos x = – \(\frac{2}{5}\), find the value of cos 2x.
Solution:
Given cos x = – \(\frac{2}{5}\)
Now we know that,
cos 2x = 2 cos2 x – 1
cos 2x = \(2\left(-\frac{2}{5}\right)^2-1\)
= \(\frac{8}{25}-1\)
= \(\frac{8-25}{25}\)
= \(-\frac{17}{25}\)

Question 3.
If tan x = \(\frac{1}{2}\) find the values of :
(i) tan 2x
(ii) sin 2x
(iii) cos 2x
Solution:
Given tan x = \(\frac{1}{2}\)

(i) tan 2x = \(\frac{2 \tan x}{1-\tan ^2 x}\)
= \(\frac{2 \times \frac{1}{2}}{1-\left(\frac{1}{2}\right)^2}\)
= \(\frac{1}{1-\frac{1}{4}}\)
= \(\frac{1}{\frac{3}{4}}=\frac{4}{3}\)

(ii) sin 2x = \(\frac{2 \tan x}{1+\tan ^2 x}\)
= \(\frac{2 \times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^2}\)
= \(\frac{1}{1+\frac{1}{4}}\)
= \(\frac{1}{\frac{5}{4}}=\frac{4}{5}\)

(iii) cos 2x = \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\)
= \(\frac{1-\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{2}\right)^2}\)
= \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}}\)
= \(\frac{\frac{3}{4}}{\frac{5}{4}}=\frac{3}{5}\)

Question 4.
If sin x = \(\frac{2}{3}\) find the value of sin 3x.
Solution:
Given sin x = \(\frac{2}{3}\)
Now, sin 3x = 3 sin x – 4 sin 3x
= \(3 \times \frac{2}{3}-4\left(\frac{2}{3}\right)^3\)
= \(2-4 \times \frac{8}{27}\)
= 2 – \(\frac{32}{27}\)
= \(\frac{54-32}{27}=\frac{22}{27}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 5.
If cos x = – \(\frac{2}{5}\), find the value of cos 3x.
Solution:
Given cos x = – \(\frac{2}{5}\)
We know that,
cos 3x = 4 cos3 x – 3 cos x
∴ cos 3x = \(4\left(-\frac{2}{5}\right)^3-3\left(-\frac{2}{5}\right)\)
= \(-\frac{32}{125}+\frac{6}{5}\)
= \(\frac{-32+150}{125}=\frac{118}{125}\)

Question 6.
Prove that :
(i) cos 2x + 2 sin2 x = 1
(ii) (cos x- sin x)2 = 1 – sin 2x
(iii) \(\frac{\cos 2 x}{\cos x-\sin x}\) = cos x + sin x
Solution:
(i) cos 2x = cos2 x – sin2 x
= 1 – sin2 x – sin2 x
= 1 – 2 sin2x
= cos2 x + 2 sin2 x = 1

(ii) LH.S. = (cos x – sin x)2
= cos2 x + sin2 x – 2 sin x cos x
= 1 – sin 2x
= R.H.S.

(iii) L.H.S. = \(\frac{\cos 2 x}{\cos x-\sin x}\)
= \(\frac{\cos ^2 x-\sin ^2 x}{\cos x-\sin x}\)
= \(\frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x-\sin x}\)
= cos x + sin x
= R.H.S.

Question 7.
Prove that :
(i) \(\frac{\sin 2 x}{1+\cos 2 x}\) = tan x
(ii) \(\frac{\sin 2 x}{1-\cos 2 x}\) = cot x
(iii) \(\frac{1-\cos 2 x}{1+\cos 2 x}\) = tan2 x
(iv) \(\frac{1+\sin 2 x-\cos 2 x}{1+\sin 2 x+\cos 2 x}\) = tan x
Solution:
(i) L.H.S. = \(\frac{\sin 2 x}{1+\cos 2 x}\)
= \(\frac{2 \sin x \cos x}{1+2 \cos ^2 x-1}\)
= \(\frac{2 \sin x \cos x}{2 \cos ^2 x}\)
= tan x
= R.H.S.

(ii) L.H.S. = \(\frac{\sin 2 x}{1-\cos 2 x}\)
= \(\frac{2 \sin x \cos x}{1-\left(1-2 \sin ^2 x\right)}\)
= \(\frac{2 \sin x \cos x}{2 \sin ^2 x}\)
= cot x
= R.H.S.

(iii) L.H.S. = \(\frac{1-\cos 2 x}{1+\cos 2 x}\)
= \(\frac{1-\left(1-2 \sin ^2 x\right)}{1+2 \cos ^2 x-1}\)
= \(\frac{2 \sin ^2 x}{2 \cos ^2 x}\)
= tan2</sup x
= R.H.S.

(iv) L.H.S. = \(\frac{1+\sin 2 x-\cos 2 x}{1+\sin 2 x+\cos 2 x}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 1

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 8.
Prove that :
(i) \(\frac{\cos ^3 x-\sin ^3 x}{\cos x-\sin x}=\frac{1}{2}\) (2 + sin 2x)
(ii) \(\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}\) = tan x
Solution:
(i) L.H.S. = \(\frac{\cos ^3 x-\sin ^3 x}{\cos x-\sin x}\)
= \(\frac{(\cos x-\sin x)\left(\cos ^2 x+\cos x \sin x+\sin ^2 x\right)}{\cos x-\sin x}\)
= cos2 x + sin2 x + sin x cos x
= 1 + \(\frac{1}{2}\) (2 sin x cos x)
= 1 + \(\frac{1}{2}\) sin 2x
= \(\frac{1}{2}\) (2 + sin 2x)
= R.H.S.

(ii) L.H.S. = \(\frac{1-\cos 2 x+\sin x}{\sin 2 x+\cos x}\)
= \(\frac{2 \sin ^2 x+\sin x}{2 \sin x \cos x+\cos x}\)
= \(\frac{\sin x(2 \sin x+1)}{\cos x(2 \sin x+1)}\)
= \(\frac{\sin x}{\cos x}\)
= tan x
= R.H.S.

Question 9.
(i) \(2 \cos 22 \frac{1}{2}^{\circ} \sin 22 \frac{1}{2}^{\circ}\)
(ii) 2 cos2 15° – 1
(iii) 8 cos3 20° – 6 cos 20°
(iv) 3 sin 40° – 4 sin3 40°
Solution:
(i) \(2 \cos 22 \frac{1}{2}^{\circ} \sin 22 \frac{1}{2}^{\circ}\)
= sin (2 × 22 \(\frac{1}{2}^{\circ}\))
= sin 45°
= \(\frac{1}{\sqrt{2}}\)

(ii) 2 cos2 15° – 1
= cos (2 × 15°)
= cos 30°
= \(\frac{\sqrt{3}}{2}\)

(iii) 8 cos3 20° – 6 cos 20° 2 [4 cos3 20° – 3 cos 20°]
= 2 cos (3 × 20°)
= 3 cos 60°
= \(\frac{3}{2}\)

(iv) 3 sin 40° – 4 sin3 40°
= sin (3 × 40°)
= sin 120°
= sin (180° – 60°)
= sin 60°
= \(\frac{\sqrt{3}}{2}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 10.
Prove that :
(i) cos \(\frac{\pi}{5}\) + cos \(\frac{3 \pi}{5}\) = \(\frac{1}{2}\)
(ii) sin2 24° – sin2 6° = \(\frac{\sqrt{5}-1}{8}\)
(iii) sin2 72° – sin2 60° = \(\frac{\sqrt{5}-1}{8}\)
(iv) sin2 72° – cos2 30° = \(\frac{\sqrt{5}-1}{8}\)
Solution:
(i) cos \(\frac{\pi}{5}\) + cos \(\frac{3 \pi}{5}\)
= cos 36° + cos 108°
= cos 36° + sin (90° + 18°)
= cos 36° – sin 18°
[∵ cos (90° + θ) = – sin θ]
= \(\frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}-1}{4}\)
= \(\frac{\sqrt{5}+1-\sqrt{5}+1}{4}\)
= \(\frac{2}{4}=\frac{1}{2}\)

(ii) sin2 24° – sin2 6° = sin (24° + 6°) sin (24° – 6°)
[∵ sin (A + B) sin (A – B) = sin2 A – sin2 B]
= sin 30° sin 18°
= \(\frac{1}{2} \times \frac{\sqrt{5}-1}{4}\)
= \(\frac{\sqrt{5}-1}{8}\)

(iii) L.H.S. = sin2 72° – sin2 60°

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 2

(iv) L.H.S. = sin2 72° – cos2 30°

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 3

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Long answer questions (11 to 26) :

Question 11.
Prove that cos x cos 2x cos 4x cos 8x = \(\frac{\sin 16 x}{16 \sin x}\)
Solution:
L.HS.= cos x cos 2x cos 4x cos 8x
= \(\frac{1}{2 \sin x}\) [2sin x cos x] cos 2x cos 4x cos 8x
= \(\frac{1}{2 \sin x}\) (sin 2x cos 2x) cos 4x cos 8x
= \(\frac{1}{2^2 \sin x}\) (2 sin 2x cos 2x) cos 4x cos 8x
= \(\frac{1}{4 \sin x}\) (sin 4x cos 4x) cos Sx
= \(\frac{1}{8 \sin x}\) (2 sin 4x cos 4x) cos 8x
= \(\frac{1}{8 \sin x}\) (sin 8x cos 8x)
[∵ sin 2θ = 2 sin θ cos θ]
= \(\frac{1}{16 \sin 2 x}\) (2 sin 8x cos 8x)
= \(\frac{\sin 16 x}{16 \sin x}\)
= R.H.S.

Question 12.
Prove that \(\frac{\cos x+\sin x}{\cos x-\sin x}-\frac{\cos x-\sin x}{\cos x+\sin x}\) = 2 tan 2x.
Solution:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 4

Question 13.
Prove that : 1 + cos2 2x = 2 (cos4 x + sin4 x).
Solution:
L.H.S. = 1 + cos2 2x
= 1 + (cos 2x)2
= 1 + (cos2 x – sin2 x)2
= (cos2 x + sin2 x)2 + (cos2 x – sin2 x)2
= cos4 x + sin4 x + 2 sin2 x cos2 x + cos4 x + sin4 x – 2 sin2 x cos2 x
= 2 (cos4 x + sin4 x)
= R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 14.
Prove that :
(i) cot \(\frac{x}{2}\) – tan \(\frac{x}{2}\) = 2 cot x
(ii) \(\sqrt{\frac{1+\sin x}{1-\sin x}}\) = tan (\(\left(\frac{\pi}{4}+\frac{x}{2}\right)\))
(iii) tan (\(\left(\frac{\pi}{4}+\frac{x}{2}\right)\)) = tan x + sec x
(iv) \(\frac{\cos 2 x}{1+\sin 2 x}\)
Solution:
(i) L.H.S. = cot \(\frac{x}{2}\) – tan \(\frac{x}{2}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 5

(ii)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 6

(iii) By using (ii),

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 7

(iv) L.H.S. = \(\frac{\cos 2 x}{1+\sin 2 x}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 8

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 15.
Prove that tan x + cot x = 2 cosec 2x and deduce that tan 75° + cot 75° = 4.
Solution:
L.H.S. = tan x + cot x
= \(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\)
= \(\frac{\sin ^2 x+\cos ^2 x}{\sin x \cos x}\)
= \(\frac{1}{\sin x \cos x}\)
= \(\frac{2}{2 \sin x \cos x}\)
= \(\frac{2}{\sin 2 x}\)
= 2 cosec 2x
= R.H.S.
Thus, tan x + cot x = 2 cosec 2x …………………(1)
putting x = 75° in eqn. (1) ; we have
tan 75° + cot 75° = 2 cosec (2 × 750)
= 2 cosec (150°)
= 2 cosec (180° – 30°)
= 2 cosec 30°
= 2 × 2 = 4

Question 16.
Prove that :
(i) tan x + \(\tan \left(\frac{\pi}{3}+x\right)+\tan \left(\frac{2 \pi}{3}+x\right)\) = 3 tan 3x
(ii) cos x \(\cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)\) = \(\frac{1}{4}\) cos 3x.
Solution:
(i) L.H.S. = tan x + \(\tan \left(\frac{\pi}{3}+x\right)+\tan \left(\frac{2 \pi}{3}+x\right)\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 9

(ii) L.H.S. = cos x \(\cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right)\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 10

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 17.
(i) If \(\frac{\sin x}{a}=\frac{\cos x}{b}\), prove that a sin 2x + b cos 2x = b.
(ii) If tan2 x = 2 tan2 y + 1, show that cos 2x + sin2 y = 0.
Solution:
Given \(\frac{\sin x}{a}=\frac{\cos x}{b}\)
⇒ tan x = \(\frac{a}{b}\) …………………(1)
L.H.S. = a sin 2x + b cos 2x

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 11

(ii) Given tan2 x = 2 tan2 y + 1 ………………(1)
L.H.S. = cos 2x + sin2 y

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 12

Question 18.
If 2 cos y = x + \(\frac{1}{x}\), prove that cos 2y = \(\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)\).
Solution:
Given 2 cos y = x + \(\frac{1}{x}\) ………………….(1)
L.H.S. = cos 2y
= 2 cos2 y – 1

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 13

Question 19.
If tan y = 3 tan x, prove that tan (x + y) = \(\frac{2 \sin 2 y}{1+2 \cos 2 y}\).
Solution:
Given tan y = 3 tan x …………………….(1)
L.H.S. = tan (x + y)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 14

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 20.
Prove that :
(i) cos 4x = 1 – 8 cos 2x + 8 cos 4x
(ii) sin 5x = 5 sin x – 20 sin3 x + 16 sin5 x.
Solution:
(i) L.H.S. = cos 4x
= cos (2 × 2x)
= 2 cos2 2x – 1
[∵ cos 2θ = 2 cos2 θ – 1]
= 2 [(2 cos2 x – 1)2]
= 2 [4 cos4 x ± 1 – 4 cos2 x] – 1
= 8 cos4 x + 2 – 8 cos2 x – 1
= 8 cos4 x – 8 cos2 x + 1
= R.H.S.

(ii) L.H.S. = sin 5x
= sin (2x + 3x)
= sin 2x cos 3x + cos 2x sin 3x
= 2 sin x cos x (4 cos3 x – 3 cos x) + (1 – 2 sin2 x) (3 sin x – 4 sin3 x)
= 2 sin x cos2 x (4 cos2 x – 3) + (1 – 2 sin2 x) (3 sin x – 4 sin3 x)
= 2 sin x (1 – sin2 x) [4 (1 – sin2 x) – 3] + (1 – 2 sin2 x) (3 sin x-4 sin3 x)
= 2 sin x (1 – sin2 x) (1 – 4 sin2 x) + (1 – 2 sin2 x) (3 sin x – 4 sin3 x)
= 2 sin x (1 – 5 sin2 x + 4sin4 x) + (3 sin x – 10 sin3 x + 8 sin5 x)
= 5 sin x – 20 sin3 x + 16 sin5 x
= R.H.S.

Question 21.
Prove that cos3 x + cos3 (\(\frac{2 \pi}{3}\) + x) + cos3 (\(\frac{2 \pi}{3}\) – x) = \(\frac{3}{4}\) cos 3x.
Solution:
L.H.S. = cos3 x + cos3 (\(\frac{2 \pi}{3}\) + x) + cos3 (\(\frac{2 \pi}{3}\) – x)
[since cos 3x = 4 cos3 x – 3 cos x
⇒ cos3 x = \(\frac{1}{4}\) (cos 3x + 3 cos x)]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 15

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 22.
Prove that :
(i) sin 6° sin 42° sin 66° sin 78° = \(\frac{1}{16}\)
(ii) \(\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}=-\frac{1}{16}\)
Solution:
(i) L.H.S. = sin 6° sin 42° sin 66° sin 78°
= \(\frac{1}{4}\) (2 sin 66° sin 6°) (2 sin 78° sin 42°)
= \(\frac{1}{4}\) [cas (66° – 6°) – cos (66° + 6°)] [cos (78° – 42°) – cos (78° + 42°)]
= \(\frac{1}{4}\) [cos 60° – cos 72°] [cos 36° – cos 120°]
= \(\frac{1}{4}\) [\(\frac{1}{2}\) – cos (90° – 18°)] [cos 36° – cos (180° – 60°)]
= \(\frac{1}{4}\) [\(\frac{1}{2}\) – sin 18°] [cos 36° + cos 60°]
= \(\frac{1}{4}\left[\frac{1}{2}-\frac{\sqrt{5}-1}{4}\right]\left[\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right]\)
= \(\frac{1}{4}\left[\frac{2-\sqrt{5}+1}{4}\right]\left[\frac{\sqrt{5}+1+2}{4}\right]\)
= \(\frac{1}{4}\left[\frac{3-\sqrt{5}}{4}\right]\left[\frac{3+\sqrt{5}}{4}\right]\)
= \(\frac{1}{64}\) (9 – 5)
= \(\frac{4}{64}=\frac{1}{16}\)
= R.H.S.

(ii) L.H.S. = \(\cos \frac{\pi}{5} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} \cos \frac{8 \pi}{5}\)
= = cos 36° cos 72° cos 144° cos 288°
[∵ \(\frac{\pi}{5}\) = 36°]
= cos 36° cos (90° – 18°) cos (180° – 36°) cos (270° + 1 8°)
= cos 36° sin 18° (- cos 36°) (sin 18°)
= – cos2 36° sin2 18°
= \(-\left(\frac{\sqrt{5}+1}{4}\right)^2\left(\frac{\sqrt{5}-1}{4}\right)^2\)
= – \(\frac{[(\sqrt{5}+1)(\sqrt{5}-1)]^2}{16 \times 16}\)
= – \(\frac{(5-1)^2}{16 \times 16}\)
= – \(\frac{1}{16}\)
= R.H.S.

Aliter:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 16

Question 23.
Given that cos \(\frac{x}{2}=\frac{12}{13}\) and x lies in first quadrant, calculate without the use of tables, the values of sin x, cos x and tan x.
Solution:
Given cos \(\frac{x}{2}=\frac{12}{13}\)
We know that, sin2 \(\frac{x}{2}\) + cos2 \(\frac{x}{2}\) = 1
⇒ sin \(\frac{x}{2}\) = ± \(\sqrt{1-\cos ^2 \frac{x}{2}}\)
⇒ sin \(\frac{x}{2}\) = ± \(\sqrt{1-\frac{144}{169}}= \pm \frac{5}{13}\)
Since x lies in first quadrant.
∴ \(\frac{x}{2}\) also lies in first quadrant
∴ sin \(\frac{x}{2}\) > 0
Thus, sin \(\frac{x}{2}=\frac{5}{13}\)
Now sin x = 2 sin \(\frac{x}{2}\) cos \(\frac{x}{2}\)
= 2 × \(\frac{5}{13} \times \frac{12}{13}=\frac{120}{169}\)
and cos x = 2 cos2 \(\frac{x}{2}\) – 1
= 2 \(\left(\frac{12}{13}\right)^2\) – 1
= \(\frac{288}{169}\) – 1
= \(\frac{119}{169}\)
Thus, tan x = \(\frac{\sin x}{\cos x}\)
= \(\frac{\frac{120}{169}}{\frac{119}{169}}=\frac{120}{119}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 24.
Given that tan x = \(\frac{12}{5}\), cos y = \(-\frac{3}{5}\) and x, y are in the same quadrant, calculate without the use of tables the values of
(i) sin (x + y)
(ii) cos \(\frac{y}{2}\)
Solution:
(i) Given tan x= \(\frac{12}{5}\), which is positive.
and cos y = \(-\frac{3}{5}\), which is negative
Thus x and y lies in third quadrant [since x andy lies in same quadrant]
We know that,
sec2 x = 1 + tan2 x
∴ sec2 x = 1 + \(\frac{144}{25}=\frac{169}{25}\)
⇒ sec x = ± \(\frac{13}{5}\)
since x lies in 3rd quadrant
∴ sec x < 0
Thus, sec x = – \(\frac{13}{5}\)
⇒ cos x = – \(\frac{5}{13}\)
∴ sin x = tan x . cos x
= \(\frac{12}{5} \times-\frac{5}{13}=-\frac{12}{13}\)
Thus sin y = – \(\sqrt{1-\cos ^2 y}\)
= – \(\sqrt{1-\frac{9}{25}}=-\frac{4}{5}\)
[∵ y lies in 3rd quad.
∴ sin y < 0]

(ii) sin (x + y) = sin x cos y + cos x sin y
= \(\left(-\frac{12}{13}\right)\left(-\frac{3}{5}\right)+\left(-\frac{5}{13}\right)\left(-\frac{4}{5}\right)\)
= \(\frac{36}{65}+\frac{20}{65}=\frac{56}{65}\)

(iii) since y lies in third quadrant thus \(\frac{y}{2}\) lies in 2nd quadrant
∴ cos \(\frac{y}{2}\) < 0
Now cos \(\frac{y}{2}\) = – \(\sqrt{\frac{1+\cos y}{2}}\)
= – \(\sqrt{\frac{1+\left(-\frac{3}{5}\right)}{2}}\)
= \(-\sqrt{\frac{1}{5}}=-\frac{1}{\sqrt{5}}\)

Question 25.
Find sin \(\frac{x}{2}\), cos \(\frac{x}{2}\) and tan \(\frac{x}{2}\) if
(i) tan x = – \(\frac{4}{3}\), x lies in second quadrant
(ii) cos x = – \(\frac{1}{3}\) x lies in third quadrant
Solution:
(i) Since 90° ≤ x ≤ 180° 45° ≤ \(\frac{x}{2}\) ≤ 90°
Thus \(\frac{x}{2}\) lies in first quadrant.
∴ sin \(\frac{x}{2}\), cos \(\frac{x}{2}\), tan \(\frac{x}{2}\) > 0
Given tan x = – \(\frac{4}{3}\)
since sec2 x = 1 + tan2 x
sec2 x = 1 + \(\frac{16}{25}\)
= \(\frac{25}{9}\)
⇒ sec x = ± \(\frac{5}{3}\)
since x lies in 2nd quadrant
∴ sec x < 0
Thus, sec x = – \(\frac{5}{3}\)
⇒ cos x = – = ± \(\frac{3}{5}\)
∴ sin x = tan x cos x
= \(-\frac{4}{3} \times\left(-\frac{3}{5}\right)=\frac{4}{5}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 17

(ii) since x lies in 3rd quadrant
∴ sin x < 0

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 18

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7

Question 26.
If tan \(\frac{x}{2}=\sqrt{\frac{a-b}{a+b}} \tan \frac{\alpha}{2}\), prove that cos x = \(\frac{a \cos \alpha+b}{a+b \cos \alpha}\).
Solution:
Given tan \(\frac{x}{2}=\sqrt{\frac{a-b}{a+b}} \tan \frac{\alpha}{2}\) …………………….(1)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.7 19

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Utilizing ISC Mathematics Class 11 Solutions Chapter 3 Trigonometry Ex 3.6 as a study aid can enhance exam preparation.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Very short answer/objective questions (1 to 3) :

Question 1.
Convert the following products into sums or differences:
(i) 2 sin 3x cos 2x
(ii) 2 cos 3x sin 2x
(iii) 2 sin 4x sin 2x
(iv) 2 cos 7x cos 3x.
Solution:
(i) 2 sin 3x cos 2x
= sin (3x + 2x) + sin (3x – 2x)
= sin 5x + sin x
[∵ 2 sin A cos B = sin (A + B) + sin (A – B)]

(ii) 2 cos 3x sin 2x
= sin (3x + 2x) – sin (3x – 2x)
= sin 5x – sin x
[∵ 2 cos A sin B = sin(A + B) – sin (A – B)]

(iii) 2 sin 4x sin 2x
= cos (4x – 2x) – cos (4x + 2x)
= cos 2x – cos 6x
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]

(iv) 2 cos 7x cos 3x
= cos (7x + 3x) + cos (7x – 3x)
= cos 10x + cos 4x
[∵ 2 cos A cos B = cos (A + B) + cos(A – B)]

Question 2.
Express each of the following as the product of sines and cosines :
(i) sin 10x + sin 6x
(ii) sin 10x – sin 6x
(ill) cos 10x + cos 6x
(iv) cos 10x – cos 6x.
Solution:
(i) sin 10x + sin 6x
= 2 sin \(\left(\frac{10 x+6 x}{2}\right)\) + cos \(\left(\frac{10 x-6 x}{2}\right)\)
= 2 sin 8x cos 2x
[∵ sin C + sin D = 2 sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]

(ii) sin 10x – sin 6x
= 2 cos \(\left(\frac{10 x+6 x}{2}\right)\) sin \(\left(\frac{10 x-6 x}{2}\right)\)
= 2 cos 8x sin 2x
[∵ sin C + sin D = 2 sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]

(iii) cos 10x + cos 6x
= – 2 sin \(\left(\frac{10 x+6 x}{2}\right)\) sin \(\left(\frac{10 x-6 x}{2}\right)\)
= – 2 sin 8x sin 2x
[∵ cos C – cos D = – 2 sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]

(iv) cos 10x – cos 6x
= – 2 sin \(\left(\frac{10 x+6 x}{2}\right)\) sin \(\left(\frac{10 x-6 x}{2}\right)\)
= – 2 sin 8x sin 2x
[∵ cos C – cos D = – 2 sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 3.
Prove that :
(i) 2 cos 45° cos 15° = \(\frac{\sqrt{3}+1}{2}\)
(ii) 2 sin 75° sin 15° = \(\frac{1}{2}\)
Solution:
(i) 2 cos 45° cos 15° = cos (45° + 15°) + cos (45° – 15°)
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= cos 60° + cos 30°
= \(\frac{1}{2}+\frac{\sqrt{3}}{2}\)
= \(\frac{1+\sqrt{3}}{2}\)

(ii) 2 sin 75° sin 15°
= cos (75° – 15°) – cos (75° + 15°)
[∵ 2 sin A sin B cos (A – B) – cos (A + B)]
= cos 60° – cos 90°
= \(\frac{1}{2}\)

Short answer questions (4 to 7) :

Question 4.
Prove that :
(i) sin 80° – cos 70° = cos 50°
(ii) cos 5° – sin 25° = sin 35°
(iii) sin 36° + cos 36° = cos 9°
(iv) cos 15°-sin 15° =7
Solution:
(i) L.H.S. = sin 80° – cos 70°
= sin 80° – cos (90° – 20°)
= sin 80° – sin 20°
= 2 cos \(\left(\frac{80^{\circ}+20^{\circ}}{2}\right)\) sin \(\left(\frac{80^{\circ}-20^{\circ}}{2}\right)\)
[∵ sin C – sin D = 2 cos \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]
= 2 cos 50° sin 30°
= 2 cos 50° × \(\frac{1}{2}\)
= cos 50°
= R.H.S.

(ii) cos 5° – sin 25°
= cos 5° – sin (90° – 65°)
= cos 5° – cos 65°
= 2 sin \(\left(\frac{5^{\circ}+65^{\circ}}{2}\right)\) sin \(\left(\frac{65^{\circ}-5^{\circ}}{2}\right)\)
[∵ cos C – cos D = 2 sin \(\frac{C+D}{2}\) sin \(\frac{C-D}{2}\)]
= 2 sin 35° × \(\frac{1}{2}\)
= sin 35°

(iii) sin 36° + cos 36°
= sin (90° – 54°) + cos 36°
= cos 54° + cos 36°
= 2 cos \(\left(\frac{54^{\circ}+36^{\circ}}{2}\right)\) cos \(\left(\frac{54^{\circ}-36^{\circ}}{2}\right)\)
[∵ cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= 2 cos 45° cos 9°
= \(\frac{2}{\sqrt{2}}\) cos 9°
= √2 cos 9°

(iv) cos 15° – sin 15°
= cos 15° – sin (90° – 75°)
= cos 15° – cos 75°
= 2 sin \(\left(\frac{15^{\circ}+75^{\circ}}{2}\right)\) sin \(\left(\frac{75^{\circ}-15^{\circ}}{2}\right)\)
= 2 sin 45° sin 30°
= 2 × \(\frac{1}{\sqrt{2}} \times \frac{1}{2}\)
= \(\frac{1}{\sqrt{2}}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 5.
Prove that :
(i) \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\) = tan 4x
(ii) \(\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}\) = cot x
(iii) \(\frac{\sin x+\sin y}{\cos x+\cos y}=\tan \frac{x+y}{2}\)
(iv) \(\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}\)
Solution:
(i) L.H.S. = \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}\)
= \(\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}\)
[using C – D formulae]
= \(\frac{2 \sin 4 x \cos x}{2 \cos 4 x \cos x}\)
= tan 4x
= R.H.S.

(ii) \(\frac{\cos 7 x+\cos 5 x}{\sin 7 x-\sin 5 x}\)
= \(\frac{2 \cos \left(\frac{7 x+5 x}{2}\right) \cos \left(\frac{7 x-5 x}{2}\right)}{2 \cos \left(\frac{7 x+5 x}{2}\right) \sin \left(\frac{7 x-5 x}{2}\right)}\)
[using C – D formulae]
= \(\frac{2 \cos 6 x \cos x}{2 \cos 6 x \sin x}\)
= cot x
= R.H.S.

(iii) \(\frac{\sin x+\sin y}{\cos x+\cos y}\)
= \(\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}\)
[∵ sin C + sin D = 2 sin \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)
cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= tan \(\left(\frac{x+y}{2}\right)\)
= R.H.S.

(iv) L.H.S. = \(\frac{\sin x-\sin y}{\cos x+\cos y}\)
= \(\frac{2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}\)
= tan \(\left(\frac{x-y}{2}\right)\)
[using C – D formulae]
= R.H.S.

Question 6.
(i) \(\frac{\cos 20^{\circ}-\cos 70^{\circ}}{\sin 70^{\circ}-\sin 20^{\circ}}\) = 1
(ii) \(\frac{\sin 75^{\circ}-\sin 15^{\circ}}{\cos 75^{\circ}+\cos 15^{\circ}}=\frac{1}{\sqrt{3}}\)
Solution:
(i) L.H.S. = \(\frac{\cos 20^{\circ}-\cos 70^{\circ}}{\sin 70^{\circ}-\sin 20^{\circ}}\)
= \(\frac{2 \sin \left(\frac{20^{\circ}+70^{\circ}}{2}\right) \sin \left(\frac{70^{\circ}-20^{\circ}}{2}\right)}{2 \cos \left(\frac{70^{\circ}+20^{\circ}}{2}\right) \sin \left(\frac{70^{\circ}-20^{\circ}}{2}\right)}\)
= tan 45°
= 1
= R.H.S.

(ii) \(\frac{\sin 75^{\circ}-\sin 15^{\circ}}{\cos 75^{\circ}+\cos 15^{\circ}}\)
= \(\frac{2 \cos \left(\frac{75^{\circ}+15^{\circ}}{2}\right) \sin \left(\frac{75^{\circ}-15^{\circ}}{2}\right)}{2 \cos \left(\frac{75^{\circ}+15^{\circ}}{2}\right) \cos \left(\frac{75^{\circ}-15^{\circ}}{2}\right)}\)
= tan 30°
= \(\frac{1}{\sqrt{3}}\)
= R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 7.
Prove that :
(i) sin (\(\frac{\pi}{4}\) sin (\(\frac{\pi}{2}\) – x) = \(\frac{1}{2}\) cos 2x
(ii) sec (\(\frac{\pi}{4}\) + x) sec (\(\frac{\pi}{4}\) – x) = 2 sec 2x
(iii) \(\sin \left(\frac{5 \pi}{6}+x\right)+\sin \left(\frac{5 \pi}{6}-x\right)\) = cos x
Solution:
(i) L.H.S. = sin (\(\frac{\pi}{4}\) sin (\(\frac{\pi}{2}\) – x)
= \(\frac{1}{2}\left[2 \sin \left(\frac{\pi}{4}+x\right) \sin \left(\frac{\pi}{4}-x\right)\right]\)
= \(\frac{1}{2}\left[\cos \left(\frac{\pi}{4}+x-\frac{\pi}{4}+x\right)\right.\) – \(\left.\cos \left(\frac{\pi}{4}+x+\frac{\pi}{4}-x\right)\right]\)
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]
= \(\frac{1}{2}\) [cos 2x – cos \(\frac{\pi}{2}\)]
= \(\frac{1}{2}\) cos 2x
= R.H.S.

(ii) L.H.S. = sec (\(\frac{\pi}{4}\) + x) sec (\(\frac{\pi}{4}\) – x)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 1

(iii) L.H.S. = \(\sin \left(\frac{5 \pi}{6}+x\right)+\sin \left(\frac{5 \pi}{6}-x\right)\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 2

Question 8.
Prove that :
(i) cos 7x + cos 5x + cos 3x + cos x = 4 cos x cos 2x cos 4x
(ii) sin x + sin 2x + sin 4x + sin 5x = 4cos \(\frac{x}{2}\) cos \(\frac{3 x}{2}\) sin 3x
(iii) cos 3x + cos 5x + cos 7x + cos 15x = 4 cos 4x cos 5x cos 6x
(iv) cos x cos \(\frac{x}{2}\) – cos 3x cos \(\frac{9 x}{2}\) = sin 4x sin \(\frac{7 x}{2}\)
Solution:
(i) L.H.S. = cos 7x + cos 5x + cos 3x + cos x
= (cos 7x + cos x) + (cos 5x + cos 3x)
= 2 \(\cos \left(\frac{7 x+x}{2}\right) \cos \left(\frac{7 x-x}{2}\right)\) + 2 \(\cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)\)
= 2 cos 4x cos 3x + 2 cos 4x cos x
= 2 cos 4x (cos 3x + cos x)
= 2 cos 4x \(\left[2 \cos \left(\frac{3 x+x}{2}\right) \cos \left(\frac{3 x-x}{2}\right)\right]\)
[∵ cos C + cos D = 2 cos \(\left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\) cos \(\left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)\)]
= 2 cos 4x (2 cos 2x cos x)
= 4 cos x cos 2x cos 4x
= R.H.S.

(ii) L.H.S. = sin x + sin 2x + sin 4x + sin 5x
= (sin 5x + sin x) + sin 4x + sin 2x)
= 2 \(\sin \left(\frac{5 x+x}{2}\right) \cos \left(\frac{5 x-x}{2}\right)\) + 2 \(\sin \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)\)
= 2 sin 3x cos 2x + 2 sin 3x cos x
= 2 sin 3x (cos 2x + cos x)
= 2 sin 3x \(\left[2 \cos \left(\frac{2 x+x}{2}\right) \cos \left(\frac{2 x-x}{2}\right)\right]\)
= 4 cos \(\frac{x}{2}\) cos \(\frac{3 x}{2}\) sin 3x
= R.H.S.

(iii) L.H.S. = cos 3x + cos 5x + cos 7x + cos 15x
= (cos 15x + cos 3x) + (cos 7x + cos 5x)
= \(2 \cos \left(\frac{15 x+3 x}{2}\right) \cos \left(\frac{15 x-3 x}{2}\right)+2 \cos \left(\frac{7 x+5 x}{2}\right) \cos \left(\frac{7 x-5 x}{2}\right)\)
= 2 cos 9x cos 6x + 2 cos 6x cos x
= 2 cos 6x [cos 9x + cos x]
= 2 cos 6x \(\left[2 \cos \left(\frac{9 x+x}{2}\right) \cos \left(\frac{9 x-x}{2}\right)\right]\)
= 2 cos 6x (2 cos 5x cos 4x)
= 4 cos 4x cos 5x cos 6x
= R.H.S.

(iv) L.H.S. = cos x cos \(\frac{x}{2}\) – cos 3x cos \(\frac{9 x}{2}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 3

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 9.
Prove that :
(i) cos 52° + cos 68° + cos 172° = 0
(ii) cos 20° + cos 100° + cos 140° = 0
(iii) \(2 \sin \frac{\pi}{17} \sin \frac{11 \pi}{17}-\cos \frac{5 \pi}{17}+\cos \frac{7 \pi}{17}\) = 0
Solution:
(I) L.H.S. = cos 52° + OS 68° COS 1720
= 2 cos \(\left(\frac{52^{\circ}+68^{\circ}}{2}\right)\) cos \(\left(\frac{52^{\circ}-68^{\circ}}{2}\right)\) + cos (180° – 8°)
[cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= 2 cos 60° cos 8° – cos 8°
= 2 × \(\frac{1}{2}\) cos 8° – cos 8°
= 0
= R.H.S.

(ii) LH.S. = cos 20° + cos 100° + cos 140°
= 2 cos \(\left(\frac{20^{\circ}+100^{\circ}}{2}\right)\) cos \(\left(\frac{20^{\circ}-100^{\circ}}{2}\right)\) + cos (180° – 40°)
= 2 cos 60° cos 40° – cos 40°
= 2 × \(\frac{1}{2}\) cos 40° – cos 40°
= 0
= R.H.S.

(iii) L.H.S = \(2 \sin \frac{\pi}{17} \sin \frac{11 \pi}{17}-\cos \frac{5 \pi}{17}+\cos \frac{7 \pi}{17}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 4

Question 10.
Prove that :
(i) sin 10° sin 50° sin 70° = \(\frac{1}{8}\)
(ii) sin 10° sin 30° sin 50° sin 70° = \(\frac{1}{16}\)
(iii) sin 20° sin 40° sin 60° sin 80° = \(\frac{3}{16}\)
Solution:
(i) L.H.S. = sin 10° sin 50° sin 70°
= \(\frac{1}{2}\) sin 10° [2 sin 70° sin 50°]
= \(\frac{1}{2}\) sin 10° [cos (20°) – cos 120°]
[∵ 2 sin A sin B = cos (A – B) – cos (A + B)]
= \(\frac{1}{2}\) sin 10° [cos 20° – cos (180° – 60°)]
= \(\frac{1}{4}\) (2 cos 20° sin 10°) – \(\frac{1}{2}\) (- cos 60°) sin 10°
= \(\frac{1}{4}\) [sin (20° + 10°) – sin (20° – 10°)] + \(\frac{1}{2}\) × \(\frac{1}{2}\) sin 10°
= \(\frac{1}{4}\) [\(\frac{1}{2}\) – sin 10°] + \(\frac{1}{4}\) sin 10°
= R.H.S.

(ii) L.H.S. = sin 10° sin 30° sin 50° sin 70°
= \(\frac{1}{2}\) [sin 10° sin 50° sin 70°]
= \(\frac{1}{2 \times 2}\) sin 10° [2 sin 70° sin 50°]
= \(\frac{1}{4}\) sin 10° [cos (70° – 50°) – cos (70° + 50°)]
= \(\frac{1}{4}\) sin 10° [cos 20° – cos(180° – 60°)]
= \(\frac{1}{4}\) sin 10° [cos 20° + cos 60°]
= \(\frac{1}{4}\) cos 20° sin 10° sin 10° × \(\frac{1}{2}\)
= \(\frac{1}{2}\) [2 cos 20° sin 10°] + \(\frac{1}{8}\) sin 10°
= \(\frac{1}{8}\) [sin (20° + 10°) – sin (20° – 10°)] + \(\frac{1}{8}\) sin 10°
= \(\frac{1}{8}\left[\frac{1}{2}-\sin 10^{\circ}\right]\)
= \(\frac{1}{16}\)
= R.H.S.

(iii) L.H.S. = sin 20° sin 40° sin 60° sin 80°
= \(\frac{\sqrt{3}}{2}\) [sin 20° sin 40° sin 80°]
= \(\frac{\sqrt{3}}{2}\) sin 20 × \(\frac{1}{2}\) [2 sin 80° sin 40°]
= \(\frac{\sqrt{3}}{4}\) sin 20° [cos (80° – 40°) – cos(80° + 40°)]
[∵ 2 sin A sin B = cos (A – B) – cos(A + B)]
= \(\frac{\sqrt{3}}{4}\) sin 20° [cos 40° – cos 120°]
= \(\frac{\sqrt{3}}{4}\) sin 20° [cos 40° – cos (180° – 60°)]
= \(\frac{\sqrt{3}}{4}\) sin 20° [cos 40° + \(\frac{1}{2}\)]
= \(\frac{\sqrt{3}}{4}\) cos 40° sin 20° + \(\frac{\sqrt{3}}{8}\) sin 20°
= \(\frac{\sqrt{3}}{8}\) (2 cos 40° sin 20°) + \(\frac{\sqrt{3}}{8}\) sin 20°
= \(\frac{\sqrt{3}}{8}\) [sin (40° + 20°) – sin (40° – 20°)] + \(\frac{\sqrt{3}}{8}\) sin 20°
[∵2 cos A sin B = sin (A + B) – sin (A – B)]
= \(\frac{\sqrt{3}}{8}\) [sin 60° – sin 20°] + \(\frac{\sqrt{3}}{8}\) sin 20°
= \(\frac{\sqrt{3}}{8} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{8} \sin 20^{\circ}+\frac{\sqrt{3}}{8} \sin 20^{\circ}\)
= \(\frac{3}{16}\)
= R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 11.
Prove that :
(i) cos 20° cos 40° cos 60° cos 80° = \(\frac{1}{16}\)
(ii) cos 10° cos 30° cos 50° cos 70° = \(\frac{3}{16}\)
Solution:
(i) L.H.S. = cos 20° cos 40° cos 60° cos 80°
= \(\frac{1}{2}\) [cos 20° cos 40° cos 80°]
= \(\frac{1}{4}\) cos 20° [2 cos 80° cos 40°]
= \(\frac{1}{4}\) cos 20° [cos (80° + 40°) + cos (80° – 40°)]
[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]
= \(\frac{1}{4}\) cos 20° [cos 120° + cos 40°]
= \(\frac{1}{4}\) cos 20° [cos (180° – 60°) + cos 40°]
= \(\frac{1}{4}\) cos 20° [- cos 60° + cos 40°]
= \(\frac{1}{4}\) cos 20° [- \(\frac{1}{2}\) + cos 40°]
= \(\frac{1}{8}\) cos 20°+ \(\frac{1}{8}\) (2 cos 40° cos 20°)
= – \(\frac{1}{4}\) cos 20° + \(\frac{1}{8}\) [cos (40° + 20°) + cos (40° – 20°)]
= – \(\frac{1}{8}\) cos 20°+ \(\frac{1}{8}\) [\(\frac{1}{2}\) + cos 20°]
= – \(\frac{1}{4}\) cos 20° + \(\frac{1}{16}\) + \(\frac{1}{8}\) cos 20°
= \(\frac{1}{16}\)
= R.HS.

(ii) L.H.S. = cos 10° cos 30° cos 50° cos 70°
= \(\frac{\sqrt{3}}{2}\) cos 10° (cos 70° cos 50°)
= \(\frac{\sqrt{3}}{4}\) cos 10° (2 cos 70° cos 50°)
= \(\frac{\sqrt{3}}{4}\) cos 10° [cos (70° + 50°) + cos (70° – 50°)]
= \(\frac{\sqrt{3}}{4}\) cos 10° [cos 120° + cos 20°]
= \(\frac{\sqrt{3}}{4}\) cos 10° [cos (180° – 60°) + cos 20°]
= \(\frac{\sqrt{3}}{4}\) cos 10° [- cos 60° + cos 20°]
= \(\frac{\sqrt{3}}{4}\) cos 10° [- \(\frac{1}{2}\) + cos 20°]
= – \(\frac{\sqrt{3}}{8}\) cos 10° + \(\frac{\sqrt{3}}{8}\) (2 cos 20° cos 10°)
= – \(\frac{\sqrt{3}}{8}\) cos 10° + \(\frac{\sqrt{3}}{8}\) [cos 30° + cos 10°]
= – \(\frac{\sqrt{3}}{8}\) cos 10° + \(\frac{\sqrt{3}}{8}\) [\(\frac{\sqrt{3}}{2}\) + cos 10°]
= \(-\frac{\sqrt{3}}{8} \cos 10^{\circ}+\frac{3}{16}+\frac{\sqrt{3}}{8} \cos 10^{\circ}=\frac{3}{16}\)
= R.H.S.

Question 12.
Prove that :
tan 10° tan 50° tan 70° = tan 30°.
Solution:
L.H.S. = tan 10° tan 50° tan 70°
= \(\frac{\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ}}{\cos 10^{\circ} \cos 50^{\circ} \cos 70^{\circ}}\) …………………(1)
Now sin 10° sin 50° sin 70° = \(\frac{1}{2}\) sin 10° (2 sin 70° sin 50°)
= \(\frac{1}{2}\) sin 10° [cos 20° – cos 120°]
[∵ 2 sinA sin B = cos (A – B) – cos (A + B)]
= \(\frac{1}{2}\) sin 10° [cos 20° – cos (180° – 60°)]
= \(\frac{1}{2}\) sin 10° [cos 20° + \(\frac{1}{2}\)]
= \(\frac{1}{4}\) (2 cos 20° sin 10°) + \(\frac{1}{4}\) sin 10°
= \(\frac{1}{4}\) [sin (20° + 10°) – sin (20° – 10°)] + \(\frac{1}{4}\) sin 10°
= \(\frac{1}{4}\) [\(\frac{1}{2}\) – sin 10°] + \(\frac{1}{4}\) sin 10°
= \(\frac{1}{8}\) …………………..(2)
= cos 10° cos 50° cos 70°
= \(\frac{1}{2}\) cos 10° (2 cos 70° cos 50°)
= \(\frac{1}{2}\) cos 10° [cos 120° + cos 20°]
= \(\frac{1}{2}\) cos 10° [cos (180° – 60°) + cos 20°]
= \(\frac{1}{2}\) cos 10° [- \(\frac{1}{2}\) + cos 20°]
= – \(\frac{1}{4}\) cos 10° + \(\frac{1}{4}\) (2 cos 20° cos 10°)
= – \(\frac{1}{4}\) cos 10° + \(\frac{1}{4}\) [cos 30° + cos 10°]
= \(\frac{1}{4}\) cos 30°
= \(\frac{1}{4} \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}\) …………………..(3)
Using eqn. (2), (3) in eqn. (1) ; we have
L.H.S. = \(\frac{\frac{1}{8}}{\frac{\sqrt{3}}{8}}=\frac{1}{\sqrt{3}}\)
= tan 30°
= R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 13.
(i) cos x + cos (\(\frac{2 \pi}{3}\) – x) + cos (\(\frac{2 \pi}{3}\) + x) = 0
(ii) \(\cos \frac{\pi}{8}+\cos \frac{3 \pi}{8}+\cos \frac{5 \pi}{8}+\cos \frac{7 \pi}{8}\) = 0.
Solution:
(i) L.H.S. = cos x + cos (\(\frac{2 \pi}{3}\) – x) + cos (\(\frac{2 \pi}{3}\) + x)
= cos x + 2 \(\cos \left(\frac{\frac{2 \pi}{3}-x+\frac{2 \pi}{3}+x}{2}\right) \cos \left(\frac{\frac{2 \pi}{3}-x-\frac{2 \pi}{3}-x}{2}\right)\)
[∵ cos C + cos D = 2 cos \(\frac{C+D}{2}\) cos \(\frac{C-D}{2}\)]
= cos x + 2 cos \(\frac{2 \pi}{3}\) cos (- x)
= cos x + 2 cos (π – \(\frac{\pi}{3}\)) cos x
= cos x + 2 (- cos \(\frac{\pi}{3}\)) cos x
= cos x – 2 × (\(\frac{1}{2}\)) cos x
= cos x – cos x
= 0
= R.H.S.

(ii) L.H.S. = \(\cos \frac{\pi}{8}+\cos \frac{3 \pi}{8}+\cos \frac{5 \pi}{8}+\cos \frac{7 \pi}{8}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 5

Question 14.
Prove that :
(i) 4 sin x sin (\(\frac{\pi}{3}\) – x) sin (\(\frac{\pi}{3}\) + x) = sin 3x
(ii) 4 cos x cos (\(\frac{\pi}{3}\) – x) cos (\(\frac{\pi}{3}\) + x) = cos 3x
Solution:
(i) L.H.S. = 4 sin x sin (\(\frac{\pi}{3}\) – x) sin (\(\frac{\pi}{3}\) + x)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 6

(ii) L.H.S. = 4 cos x cos (\(\frac{\pi}{3}\) – x) cos (\(\frac{\pi}{3}\) + x)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 7

= – cos x + 2 cos 2x cos x
= – cos x + cos (2x + x) + cos (2x – x)
= – cos x + cos 3x + cos x
= cos 3x
= R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 15.
Prove that :
(i) (cos x + cosy)2 + (sin x + siny)2 = 4 c0s2
(ii) (cos x + cos y)2 + (sin x – sin y)2 = 4 cos2
(iii) sin2 x + sin2 (x – y) – 2 sin x cos y sin (x – y) = sin2 y.
Sol.
(i) L.H.S. = (cos x + cos y)2 – (sin x + sin y)2

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 8

(ii) L.H.S. = (cos x + cos y)2 + (sin x – sin y)2

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 9

(iii) L.H.S. = sin2 x + sin2 (x – y) – 2 sin x cos y sin (x – y)
= sin2 x + sin (x – y) [sin (x – y) – 2 sin x cos y]
= sin2 x + sin (x – y) [sin (x – y) – sin (x + y) – sin (x – y)]
= sin2 x + sin (x – y) {- sin (x + y)}
= sin2 x – sin (x – y) sin (x + y)
= sin2 x – {sin2 x – sin2 y}
= sin2 x – sin2 x + sin2 y
= sin2 y
= R.H.S.

Question 16.
Prove that :
(i) \(\frac{\sin x+\sin y}{\sin x-\sin y}=\tan \frac{x+y}{2} \cot \frac{x-y}{2}\)
(ii) \(\frac{\sin x+\sin 3 x+\sin 5 x}{\cos x+\cos 3 x+\cos 5 x}\) = tan 3x
(iii) \(\frac{\sin 5 x+2 \sin 8 x+\sin 11 x}{\sin 8 x+2 \sin 11 x+\sin 14 x}=\frac{\sin 8 x}{\sin 11 x}\)
(iv) \(\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\) = tan 2x
(v) \(\frac{\cos x+\cos 3 x+\cos 5 x+\cos 7 x}{\sin x+\sin 3 x+\sin 5 x+\sin 7 x}\) = cot 4x
Solution:
(i) L.H.S. = \(\frac{\sin x+\sin y}{\sin x-\sin y}\)
= \(\frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right)}\)
= \(\tan \left(\frac{x+y}{2}\right) \cot \left(\frac{x-y}{2}\right)\)
= R.H.S

(ii) L.H.S. = \(\frac{\sin x+\sin 3 x+\sin 5 x}{\cos x+\cos 3 x+\cos 5 x}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 10

(iii) L.H.S. = \(\frac{\sin 5 x+2 \sin 8 x+\sin 11 x}{\sin 8 x+2 \sin 11 x+\sin 14 x}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 11

(iv) L.H.S. = \(\frac{\sin 8 x \cos x-\sin 6 x \cos 3 x}{\cos 2 x \cos x-\sin 3 x \sin 4 x}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 12

(v) \(\frac{\cos x+\cos 3 x+\cos 5 x+\cos 7 x}{\sin x+\sin 3 x+\sin 5 x+\sin 7 x}\)
= cot4 x

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 13

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 17.
Prove that :
cos α + cos β + cos γ + cos (α + β + γ) = 4 \(\cos \frac{\alpha+\beta}{2} \cos \frac{\beta+\gamma}{2} \cos \frac{\gamma+\alpha}{2}\)
Solution:
L.H.S. = cos α + cos β) + cos γ + cos (α + β + γ)
= (cos α + cos β) + (cos (α + β + γ) + cos γ)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 14

Question 18.
If cos x + cos y = \(\frac{1}{3}\) and sin x + sin y = \(\frac{1}{4}\), prove that \(\frac{x+y}{2}=\frac{3}{4}\).
Solution:
Given cos x + cos y = \(\frac{1}{3}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 15

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6

Question 19.
(i) \(\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}\) then show that \(\frac{\tan x}{\tan y}=\frac{a}{b}\).
(ii) If cos (x + 2y) = m cos x, prove that cot y = \(\frac{1+m}{1-m}\) tan (x + y).
Solution:
Given \(\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}\)
Applying componendo and dividendo, we have

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.6 16

(ii) Given, cos (x + 2y) = m cos x
⇒ \(\frac{\cos (x+2 y)}{\cos x}=\frac{m}{1}\)
Applying componendo and dividendo, we have

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

The availability of step-by-step Understanding ISC Mathematics Class 11 Solutions Chapter 3 Trigonometry Ex 3.5 can make challenging problems more manageable.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Very short answer/objective questions (1 to 3) :

Question 1.
Find the values of :
(i) cos 210°
(ii) sin 225°
(iii) tan 330°
(iv) sin 930°
(v) cos (- 870°)
(vi) cos \(\frac{2 \pi}{3}\)
(vii) sin \(\frac{19 \pi}{4}\)
(viii) tan (- \(\frac{4 \pi}{3}\))
Solution:
(i) cos 210° = cos (180° + 30°)
= – cos 30°
= – \(\frac{\sqrt{3}}{2}\)
[∵ cos (180° + θ) = – cos θ]

(ii) sin 225° = sin (180° + 45°)
= – sin 45°
= – \(\frac{1}{\sqrt{2}}\)
[∵ sin (180° + θ) = – sin θ]

(iii) tan 330° = tan (360° – 30°)
= – tan 30°
= – \(\frac{1}{\sqrt{3}}\)
[∵ tan (360° – θ) = – tan θ]

(iv) sin 930° = sin (720° + 210°)
= sin 210°
= sin (180° + 30°)
= – sin 30°
= – \(\frac{1}{2}\)
[∵ sin (180° + θ) = – sin θ]

(v) cos (- 870°) = cos 870°
[∵ cos (- θ) = cos θ]
= cos (720° + 150°)
= cos 150°
= cos (180° – 30°)
= – cos 30°
= – \(\)
[∵ cos (180° – θ) = – cos θ]

(vi) cos \(\frac{2 \pi}{3}\) = cos (π – \(\frac{\pi}{3}\))
= – cos \(\frac{\pi}{3}\)
= – \(\frac{1}{2}\)

(vii) sin \(\frac{19 \pi}{4}\) = sin (4π + \(\frac{3 \pi}{4}\))
= sin \(\frac{3 \pi}{4}\)
= sin (π – \(\frac{\pi}{4}\))
= sin \(\frac{\pi}{4}\)
= \(\frac{1}{\sqrt{2}}\)
[∵ sin (π – θ) = sin θ]

(viii) tan (- \(\frac{4 \pi}{3}\)) = – tan (\(\frac{4 \pi}{3}\))
[∵ tan (- θ) = – tan θ]
= – taan (π + \(\frac{\pi}{3}\))
= – tan \(\frac{\pi}{3}\)
= – √3
[∵ tan (π + θ) = tan θ]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 2.
Find the values of :
(i) sin2 \(\frac{\pi}{2}\) + cos2 \(\frac{\pi}{3}\) – tan2 \(\frac{\pi}{4}\)
(ii) tan2 \(\frac{\pi}{3}\) + 2 cos2 \(\frac{\pi}{4}\) + 3 sec2 \(\frac{\pi}{6}\) – 4 cos2 \(\frac{\pi}{2}\).
Solution:
(i) sin2 \(\frac{\pi}{2}\) + cos2 \(\frac{\pi}{3}\) – tan2 \(\frac{\pi}{4}\)
= \(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2\) – 12
= \(\frac{1}{4}+\frac{1}{4}\) – 1 = – \(\frac{1}{2}\)

(ii) tan2 \(\frac{\pi}{3}\) + 2 cos2 \(\frac{\pi}{4}\) + 3 sec2 \(\frac{\pi}{6}\) – 4 cos2 \(\frac{\pi}{2}\)
= (√3)2 + 2 \(\left(\frac{1}{\sqrt{2}}\right)^2\) + 3 \(\left(\frac{2}{\sqrt{3}}\right)^2\) – 4 × 02
= 3 + 2 × \(\frac{1}{2}\) + 3 × \(\frac{4}{3}\) – 0
= 3 + 1 + 4 = 8

Question 3.
If tan x = \(\frac{2}{3}\) and tan y = \(\frac{3}{4}\), find the value of tan (x + y).
Solution:
Given tan x = \(\frac{2}{3}\) ;
tan y = \(\frac{3}{4}\)
∴ tan (x + y) = \(\frac{\tan x+\tan y}{1-\tan x \tan y}\)
= \(\frac{\frac{2}{3}+\frac{3}{4}}{1-\frac{2}{3} \times \frac{3}{4}}\)
= \(\frac{\frac{17}{12}}{\frac{1}{2}}\)
= \(\frac{34}{12}=\frac{17}{6}\)

Short answer questions (4 to 6) :

Question 4.
Express the following as functions of angles less than 45° :
(i) sin (- 1785°)
(ii) cosec (- 7498°)
Soluion:
(i) sin (- 1785°) = – sin (1 785°)
[∵ sin(- θ) = – sin θ]
= – sin (1440° + 345°)
= – sin (4 X 360° + 3450)
= – sin (345°)
= – sin (360 – 15°)
= sin 15°
[∵ sin (360° – θ) = – sin θ]
= sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
[∵ sin (A – B) = sin A cos B – cos A sin B]
= \(\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \times \frac{1}{2}\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)

(ii) cosec (- 7498°) = – cosec (7498°)
[∵ cosec (- θ) = – cosec θ]
= – cosec (360° × 20 + 298°)
– cosec (298°)
= – cosec (270° + 28°)
= sec 28°
[∵ cosec (270° + θ) – sec θ]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 5.
(i) Which is bigger: sin 55° or cos 55°?
(ii) If θ = 100°, determine the sign of (sin θ + cos θ).
Solution:
Now cos 55° = cos (90° – 35°) = sin 35°
since sin x be an increasing function in first quadrant
since 55° > 35°
⇒ sin 55° > sin 35° = cos 55°
[if x1 > x2 and f be an increasing function.
Then f (x1) > f (x2)]
Thus sin 55° is bigger.

(ii) When θ = 1000
∴ sin θ + cos θ = sn 100° + cos 100°
= sin 100° + cos (90° + 10°)
= sin 100°- sin 10°
[∵ cos (90° + θ) = – sin θ]
= sin (90° + 10°) – sin 10°
= cos 10° – sin 10° > 0
[∵ When 0 < θ < \(\frac{\pi}{4}\) ⇒ cos θ > sin θ]

Question 6.
Prove that :
(i) cot A + tan (π + A) + tan (\(\frac{\pi}{2}\) + A) + tan (2π – A) = 0
(ii) \(2 \sin ^2 \frac{\pi}{6}+\ {cosec} \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}\) = 0
(iii) \(3 \cos ^2 \frac{\pi}{4}+\sec \frac{2 \pi}{3}+5 \tan ^2 \frac{\pi}{3}=\frac{29}{2}\)
(iv) \(2 \cos ^2 \frac{\pi}{4}+2 \sin ^2 \frac{3 \pi}{4}+2 \sec ^2 \frac{\pi}{3}\) = 10
(v) \(\cot ^2 \frac{\pi}{6}+\ {cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6}\) = 6
(vi) \(\left(3 \cos \frac{\pi}{3} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \tan \frac{\pi}{4}\right)\) cos 2π = 1
Solution:
(i) L.H.S = cot A + tan (π + A) + tan (\(\frac{\pi}{2}\) + A) + tan (2π – A)
= cot A + tan A – cot A – tan A = 0
= R.H.S.

(ii) L.H.S. = \(2 \sin ^2 \frac{\pi}{6}+\ {cosec} \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}\)
= 2 (\(\frac{1}{2}\))2 + cosec (π + \(\frac{\pi}{6}\)) × (\(\frac{1}{2}\))2
= 2 × \(\frac{1}{4}\) – cosec \(\frac{\pi}{6} \times \frac{1}{4}\)
[∵ cosec (π + θ) = – cosec θ]
= \(\frac{1}{2}-2 \times \frac{1}{4}\) = 0
= R.H.S.

(iii) L.H.S. = \(3 \cos ^2 \frac{\pi}{4}+\sec \frac{2 \pi}{3}+5 \tan ^2 \frac{\pi}{3}\)
= \(3\left(\frac{1}{\sqrt{2}}\right)^2+\sec \left(\pi-\frac{\pi}{3}\right)+5(\sqrt{3})^2\)
= \(\frac{3}{2}-\sec \frac{\pi}{3}\) + 15
[∵ sec (π – θ) = – cosec θ]
= \(\frac{3}{2}\) – 2 + 15
= \(\frac{29}{2}\)
= R.H.S.

(iv) L.H.S. = \(2 \cos ^2 \frac{\pi}{4}+2 \sin ^2 \frac{3 \pi}{4}+2 \sec ^2 \frac{\pi}{3}\)
= \(2\left(\frac{1}{\sqrt{2}}\right)^2+2 \sin ^2\left(\pi-\frac{\pi}{4}\right)\) + 2 (2)2
= 2 × \(\frac{1}{2}\) + 2 \(\left(\frac{1}{\sqrt{2}}\right)^2\) + 8
= 1 + 1 + 8 = 10
= R.H.S.

(v) L.H.S. = \(\cot ^2 \frac{\pi}{6}+\ {cosec} \frac{5 \pi}{6}+3 \tan ^2 \frac{\pi}{6}\)
= \((\sqrt{3})^2+\ {cosec}\left(\pi-\frac{\pi}{6}\right)+3\left(\frac{1}{\sqrt{3}}\right)^2\)
= 3 + cosec \(\frac{\pi}{6}\) + 3 × \(\frac{1}{3}\)
[∵ cosec (π – θ) = cosec θ]
= 3 + 2 + 1 = 6
= R.H.S.

(vi) L.H.S. = \(\left(3 \cos \frac{\pi}{3} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \tan \frac{\pi}{4}\right)\) cos 2π
= [3 × \(\frac{1}{2}\) × 2 – 4 sin (π – \(\frac{\pi}{6}\)) × 1] × 1
= [3 – 4 sin \(\frac{\pi}{6}\)]
= 3 – 4 × \(\frac{1}{2}\)
= 3 – 2 = 1
= R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Long answer questions (7 to 32) :

Question 7.
Evaluate the following :
(i) √2 sin 135° cos 210° tan 240° cot 300° sec 330°
(ii) sin 690° cos 930° + tan (- 765°) cosec (- 1170°)
(iii) \(\sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8}+\sin ^2 \frac{5 \pi}{8}+\sin ^2 \frac{7 \pi}{8}\)
(iv) \(\sin ^2 \frac{\pi}{4}+\sin ^2 \frac{3 \pi}{4}+\sin ^2 \frac{5 \pi}{4}+\sin ^2 \frac{7 \pi}{4}\)
(v) \(\tan \frac{\pi}{12} \tan \frac{\pi}{16} \tan \frac{5 \pi}{12} \tan \frac{7 \pi}{16}\)
Solution:
(i) sin 135° = sin(180° – 45°)
= sin 45°
= \(\frac{1}{\sqrt{2}}\)
[∵ sin(180° – θ) = sin θ]
cos 210° = cos(180° + 30°)
= – cos 30°
= – \(\frac{\sqrt{3}}{2}\)
tan 240° = tan (180° + 60°)
= tan 60° = √3
cot 300° = cot(360° – 60°)
= – cot 60°
= \(\frac{1}{\sqrt{32}}\)
sec 330° = sec (360° – 30°)
= sec 30°
= \(\frac{2}{\sqrt{3}}\)
Thus, √2 sin 135° cos 210° tan 240° cot 300° sec 330° = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\left(-\frac{\sqrt{3}}{2}\right) \sqrt{3}\left(-\frac{1}{\sqrt{3}}\right)\left(\frac{2}{\sqrt{3}}\right)\) = 1

(ii) Now sin 690° = sin (720° – 30°)
= sin (- 30°)
= – \(\frac{1}{2}\)
cos 930° = cos (720° + 210°)
= cos 210°
= cos (180° + 30°)
= – cos30°
tan (- 765°) = – tan (765°)
= – tan (720° + 45°)
= – tan 45° = – 1
and cosec (- 1170°) = – cosec (1080° + 90°)
= – cosec 90° – 1
Now, sin 690° cos 930° + tan (- 765°) cosec (- 1170°) = \(\left(-\frac{1}{2}\right)\left(-\frac{\sqrt{3}}{2}\right)\) + (- 1) (- 1)
= \(\frac{\sqrt{3}}{4}\) + 1

(iii) \(\sin ^2 \frac{\pi}{8}+\sin ^2 \frac{3 \pi}{8}+\sin ^2 \frac{5 \pi}{8}+\sin ^2 \frac{7 \pi}{8}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 1

(iv) \(\sin ^2 \frac{\pi}{4}+\sin ^2 \frac{3 \pi}{4}+\sin ^2 \frac{5 \pi}{4}+\sin ^2 \frac{7 \pi}{4}\)
= \(\sin ^2 \frac{\pi}{4}+\sin ^2\left(\frac{\pi}{2}+\frac{\pi}{4}\right)+\sin ^2 \frac{5 \pi}{4}+\sin ^2\left(\frac{\pi}{2}+\frac{5 \pi}{4}\right)\)
= \(\sin ^2 \frac{\pi}{4}+\cos ^2 \frac{\pi}{4}+\sin ^2 \frac{5 \pi}{4}+\cos ^2 \frac{5 \pi}{4}\)
[∵ sin (\(\frac{\pi}{2}\) + θ) = cos θ]
= 1 + 1 = 2

Aliter:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 2

(v) \(\tan \frac{\pi}{12} \tan \frac{\pi}{16} \tan \frac{5 \pi}{12} \tan \frac{7 \pi}{16}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 3

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 8.
Prove that :
cos x + sin (\(\frac{3 \pi}{2}\) + x) – sin (\(\frac{3 \pi}{2}\) – x) + cos (π + x) = 0.
Solution:
L.H.S. = cos x + sin (\(\frac{3 \pi}{2}\) + x) – sin (\(\frac{3 \pi}{2}\) – x) + cos (π + x)
= cos x – cos x + cos x – cos x
= 0
= R.H.S.

Question 9.
Simplify the following:
(i) \(\frac{\cos x}{\sin \left(\frac{\pi}{2}+x\right)}+\frac{\sin (-x)}{\sin (\pi+x)}-\frac{\tan \left(\frac{\pi}{2}+x\right)}{\cot x}\)
Soluion:
\(\frac{\cos x}{\sin \left(\frac{\pi}{2}+x\right)}+\frac{\sin (-x)}{\sin (\pi+x)}-\frac{\tan \left(\frac{\pi}{2}+x\right)}{\cot x}\)
= \(\frac{\cos x}{\cos x}-\frac{\sin x}{-\sin x}-\frac{(-\cot x)}{\cot x}\)
= 1 + 1 + 1
= 3

(ii) \(\frac{\sin (\pi+x) \cos \left(\frac{\pi}{2}+x\right) \tan \left(\frac{3 \pi}{2}-x\right) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \ {cosec}(-x) \sin \left(\frac{3 \pi}{2}+x\right)}\)
Soluion:
\(\frac{\sin (\pi+x) \cos \left(\frac{\pi}{2}+x\right) \tan \left(\frac{3 \pi}{2}-x\right) \cot (2 \pi-x)}{\sin (2 \pi-x) \cos (2 \pi+x) \ {cosec}(-x) \sin \left(\frac{3 \pi}{2}+x\right)}\)
= \(\frac{(-\sin x)(-\sin x)(\cot x)(-\cot x)}{(-\sin x) \cos x(-\ {cosec} x)(-\cos x)}\)
= \(\frac{\sin ^2 x \cot ^2 x}{\cos ^2 x}\)
= tan2 x cot2 x = 1

Question 10.
Find y from the following equation:
cosec (\(\frac{\pi}{2}\) + x) + y cos x cot (\(\frac{\pi}{2}\) + x) = sin (\(\frac{\pi}{2}\) + x)
Solution:
Given,
cosec (\(\frac{\pi}{2}\) + x) + y cos x cot (\(\frac{\pi}{2}\) + x) = sin (\(\frac{\pi}{2}\) + x)
⇒ + sec x + y cos x (- tan x) = cos x
⇒ sec x – cos x = y cos x \(\left(\frac{\sin x}{\cos x}\right)\)
⇒ \(\frac{1}{\ cos x}\) – cos x = y sin x
⇒ \(\frac{1-\cos ^2 x}{\cos x}\) = y sin x
⇒ \(\frac{\sin ^2 x}{\cos x}\) = y sin x
⇒ y = tan x

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 11.
If 8x = π, show that cos 7x + cos x = 0.
Solution:
Given 8x = π
L.H.S. = cos 7x + cos x
= cos \(\frac{7 \pi}{8}\) + cos \(\frac{\pi}{8}\)
= cos (π – \(\frac{\pi}{8}\)) + cos \(\frac{\pi}{8}\)
= – cos \(\frac{\pi}{8}\) + cos \(\frac{\pi}{8}\)
= 0
= R.H.S.
= \(-\frac{1}{\tan 25^{\circ}}=-\frac{1}{x}\)
[∵ cos (π – θ) = – cos θ]

Question 12.
If tan 25° = x, prove that \(\frac{\tan 155^{\circ}-\tan 115^{\circ}}{1+\tan 155^{\circ} \tan 115^{\circ}}=\frac{1-x^2}{2 x}\)
Solution:
Now, tan 155° = tan (180° – 25°)
= – tan 25° = – x
[∵ tan 25° = x]
and tan 115° = tan(90° + 25°)
= – cot 25°
= \(-\frac{1}{\tan 25^{\circ}}=\frac{-1}{x}\)
LH.S. = \(\frac{\tan 155^{\circ}-\tan 115^{\circ}}{1+\tan 155^{\circ} \tan 115^{\circ}}\)
= \(\frac{-x-\left(-\frac{1}{x}\right)}{1+(-x)\left(-\frac{1}{x}\right)}\)
= \(\frac{-x+\frac{1}{x}}{1+1}\)
= \(\frac{1-x^2}{2 x}\)
= R.H.S.

Question 13.
Prove that \(\frac{\cos 135^{\circ}-\cos 120^{\circ}}{\cos 135^{\circ}+\cos 120^{\circ}}\) = 3 – 2√2
Solution:
L.H.S. = \(\frac{\cos 135^{\circ}-\cos 120^{\circ}}{\cos 135^{\circ}+\cos 120^{\circ}}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 4

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 14.
If A, B, C are angles of a triangle, prove that
(i) sin (A + B) = sin C
(ii) cos \(\frac{A+B}{2}\) = sin \(\frac{C}{2}\)
(iii) \(\frac{\tan (B+C)+\tan (C+A)+\tan (A+B)}{\tan (\pi-A)+\tan (2 \pi-B)+\tan (3 \pi-C)}\) = 1
Solution:
Since A, B, C are the angles of a triangle
∴ A + B + C = π
⇒ A + B = π – C

(i) sin (A + B) = sin (π – C)
= sin C

(ii) cos \(\frac{A+B}{2}\) = cos \(\left(\frac{\pi-C}{2}\right)\)
= cos \(\left(\frac{\pi}{2}-\frac{C}{2}\right)\)
= sin \(\frac{C}{2}\)

(iii) \(\frac{\tan (B+C)+\tan (C+A)+\tan (A+B)}{\tan (\pi-A)+\tan (2 \pi-B)+\tan (3 \pi-C)}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 5

Question 15.
If A, B, C, D are angles of a quadrilateral, then prove that
(i) sin (A + D) + sin (B + C) = 0
(ii) \(\sin \left(\frac{A+B}{2}\right)=\sin \left(\frac{C+D}{2}\right)\)
(iii) \(\tan \left(\frac{A+B}{2}\right)+\tan \left(\frac{C+D}{2}\right)\) = 0
(iv) cosec2 (A + B) = cosec2 (C + D)
(v) sin A + sin (B + C + D ) = 0.
Solution:
(i) Since A, B, C and D are the angles of a quadrilateral.
∴ A + B + C + D = 360°
⇒ A + D = 360° – (B + C)
⇒ sin (A + D) = sin \(\left[360^{\circ}-\overline{\mathrm{B}+\mathrm{C}}\right]\)
= – sin (B + C)
sin (A + D) + sin (B + C) = 0

(ii) A + B + C + D = 360°

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 6

(iii) ∴ tan \(\left(\frac{A+B}{2}\right)\)
= tan [180° – \(\left(\frac{C+D}{2}\right)\)] [using (1)]
= tan \(\left(\frac{C+D}{2}\right)\)
[∵ tan (180 – θ) = – tan θ]
⇒ \(\tan \left(\frac{A+B}{2}\right)+\tan \left(\frac{C+D}{2}\right)\) = 0

(iv) since A + B + C + D = 360°
⇒ A + B = 360° – (C +D)
⇒ cosec (A + B) = cosec2 [360° – (C + D)]
= [cosec2 (360° – \(\overline{C+D}\))]2
= [- cosec (C + D)]2
= cosec2 (C + D)

(v) since A + B + C + D = 2π
⇒ B + C + D = 2π – A
⇒ sin (B + C + D) = sin (2π – A)
= – sin A
⇒ sin A + sin (B + C + D) = 0

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 16.
If A, B, C, D are angles of a cyclic quadrilateral, prove that
(i) cot A + cot B + cot C + cot D = 0
(ii) sin A + sin B = sin C + sin D.
Solution:
Since A, B, C and D are the angles of cyclic quadrilateral.
∴ A + C = B + D = π ………………..(1)

(i) L.H.S. = cot A + cot B + cot C + cot D
= cot A + cot B + cot (n – A) + cot (n – B) [using (1)]
= cot A + cot B – cot A – cot B = 0
= R.H.S.

(ii) L.H.S. = sin A + sin B
= sin (π – C) + sin (π – D) [using (1)]
= sin C + sin D
= R.H.S.
[∵ sin (π – θ) = sin θ]

Question 17.
Find tan 15° and hence show that tan 15° + cot 15° = 4.
Solution:
tan 150° = tan (45° – 30°)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 7

Question 18.
Evaluate:
(i) cos 195°
(ii) sin \(\frac{11 \pi}{12}\)
Solution:
(i) cos 195° = cos (180° + 15°)
= – cos 15°
[∵ cos (180° – θ) = – cos θ]
= – cos (45° – 30°)
= – [cos 45° cos 30° + sin 45° sin 30°]
[∵ cos (A – B) = cos A cos B + sin A sin B]
= – \(\left[\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}\right]\)
= – \(\left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right)\)

(ii) sin \(\frac{11 \pi}{12}\)
= sin (π \(\frac{\pi}{12}\))
= sin \(\frac{\pi}{12}\)
=sin 15°
= sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
[∵ sin (A – B) = sin A cos B – cos A sin B]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 19.
Evaluate the following:
(i) cos 105° + sin 105°
(ii) cos 15° – sin 15°
(iii) cot 105° – tan 105°.
Solution:
(i) cos 105° + sin 105°
= cos (60° + 45°) + sin (60° + 45°)
= [cos 64° cos 45° – sin 60° sin 45°] + [sin 60° cos 45° + cos 60° sin 45°]
= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2} \frac{1}{\sqrt{2}}+\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}\)

(ii) cos 15° = cos (45° – 30°)
= cos 45° cos 30° + sin 45° sin 30°
[∵ cos (A – B) = cos A cos B + sin A sin B]
= \(\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \frac{1}{2}\)
= \(\frac{\sqrt{3}+1}{2 \sqrt{2}}\)
sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
[∵ sin (A – B) = sin A cos B – cos A sin B]
= \(\frac{1}{\sqrt{2}} \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \frac{1}{2}\)
= \(\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
∴ cos 15° – sin 15° = \(\frac{\sqrt{3}+1}{2 \sqrt{2}}-\frac{\sqrt{3}-1}{2 \sqrt{2}}\)
= \(\frac{\sqrt{3}+1-\sqrt{3}+1}{2 \sqrt{2}}\)
= \(\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}\)

(iii) cot 105° – tan 105° = cot (90° + 15°) – tan 105°
= – tan 15° – tan 105°
= – tan (45° – 30°) – tan (60° + 45°)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 8

Question 20.
Show that:
(i) sin 38° cos 22° + cos 38° sin 22° = \(\frac{\sqrt{3}}{2}\)
(ii) sin 70° cos 10° – cos 70° sin 10° = \(\frac{\sqrt{3}}{2}\)
(iii) cos 130° cos 40° + sin 130° sin 40° = 0
(iv) \(\sin \frac{7 \pi}{12} \cos \frac{\pi}{4}-\cos \frac{7 \pi}{12} \sin \frac{\pi}{4}=\frac{\sqrt{3}}{2}\)
(v) \(\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}\) = – 1
Solution:
(i) sin 38° cos 22° + cos 38° sin 22°
= sin (38° + 22°)
= sin 60°
= \(\frac{\sqrt{3}}{2}\)
[∵ sin (A – B) = sin A co sB – cos A sin B]

(ii) sin 70° cos 10° – cos 70° sin 10° = sin (70° – 10°)
= sin 60°
= \(\frac{\sqrt{3}}{2}\)
[∵ sin (A – B) = sin A cos B + cos A sin B

(iii) cos 130° cos40° + sin 130° sin 40°
= cos (130° – 40°)
= cos 90° = 0
[∵ cos (A – B) = cos A cos B + sin A sin B]

(iv) \(\sin \frac{7 \pi}{12} \cos \frac{\pi}{4}-\cos \frac{7 \pi}{12} \sin \frac{\pi}{4}\)
= sin \(\left(\frac{7 \pi}{12}-\frac{\pi}{4}\right)\)
= sin \(\left(\frac{4 \pi}{12}\right)\)
= sin \(\frac{\pi}{3}\)
= \(\frac{\sqrt{3}}{2}\)
[∵ sin A cos B – cos A sin B = sin (A – B)]

(v) \(\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}\) = tan (69° + 66°)
= tan 135°
= tan (90° + 45°)
= – cot 45° = – 1
[∵ tan (A + B) = \(\frac{\tan A+\tan B}{1-\tan A \tan B}\)]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 21.
Prove that :
(i) sin (x – y) cos x – cos (x – y) sin x = – sin y
(ii) √2 cos (\(\frac{\pi}{4}\) + x) = cos x – sin x.
Solution:
(i) sin (x – y) cos x – cos (x – y) sin x = sin (x – y – x)
= sin (- y)
= – sin y
[∵ sin A cos B – cos A sin B = sin (A – B)]

(ii) L.H.S. = √2 cos (\(\frac{\pi}{4}\) + x)
= √2 [cos \(\frac{\pi}{4}\) cos x – sin \(\frac{\pi}{4}\) sin x]
[∵ cos (A + B) = cos A cos B – sin A sin B]
= √2 [\(\frac{1}{\sqrt{2}}\) cos x – \(\frac{1}{\sqrt{2}}\) sin x]
= cos x – sin x
= R.H.S.

Question 22.
(i) \(\cos ^2\left(\frac{\pi}{4}+x\right)-\sin ^2\left(\frac{\pi}{4}-x\right)\)
(ii) \(\sin ^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin ^2\left(\frac{\pi}{8}-\frac{x}{2}\right)\)
Soluion:
(i) \(\cos ^2\left(\frac{\pi}{4}+x\right)-\sin ^2\left(\frac{\pi}{4}-x\right)\)
= cos (\(\frac{\pi}{4}\) + x + \(\frac{\pi}{2}\) – x) cos (\(\frac{\pi}{2}\) + x – \(\frac{\pi}{2}\) + x)
[∵ cos2 A – sin2 B = cos (A + B) – cos (A – B)]
= cos \(\frac{\pi}{2}\) cos 2x = 0

(ii) \(\sin ^2\left(\frac{\pi}{8}+\frac{x}{2}\right)-\sin ^2\left(\frac{\pi}{8}-\frac{x}{2}\right)\)
= \(\sin \left(\frac{\pi}{8}+\frac{x}{2}+\frac{\pi}{8}-\frac{x}{2}\right) \sin \left(\frac{\pi}{8}+\frac{x}{2}-\frac{\pi}{8}+\frac{x}{2}\right)\)
[∵ sin2 A – sin2 B = sin (A + B) – sin (A – B)]
= sin \(\frac{\pi}{4}\) sin x
= \(\frac{1}{\sqrt{2}}\) sin x

Question 23.
(i) If sin A = \(\frac{1}{\sqrt{5}}\) and cos B = \(\frac{3}{\sqrt{10}}\), where A, B are positive acute angles, prove that A + B = 45°.
(ii) If cos A = \(\frac{1}{7}\) and cos B = \(\frac{13}{14}\) and A, B lie in first quadrant, prove that A – B = 60°.
Solution:
(i) Given sin A = \(\frac{1}{\sqrt{5}}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 9

(ii) Given cos A = \(\frac{1}{7}\)
and cos B = \(\frac{13}{14}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 10

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 24.
(i) If α, β lie in first quadrant and sin α = \(\frac{8}{17}\), tan β = \(\frac{5}{12}\), find the values of sin (α – β), cos (α – β) and tan (α – β).
(ii) If cos x = \(\frac{4}{5}\), cos y = \(\frac{12}{13}\), \(\frac{3 \pi}{2}\) < x < 2π and \(\frac{3 \pi}{2}\) < y < 2π, find the values of cos (x +y) and sin (x – y).
Solution:
Given sin α = \(\frac{8}{17}\)
and tan β = \(\frac{5}{12}\)
∴ cos α = \(\sqrt{1-\sin ^2 \alpha}\)
= \(\sqrt{1-\frac{64}{289}}\)
= \(\sqrt{\frac{225}{289}}=\frac{15}{17}\)
[∵ α lies in first quadrant]
and sec2 β = 1 + tan2 β
= 1 + \(\left(\frac{5}{12}\right)^2\)
= 1 + \(\frac{25}{144}\)
= \(\frac{169}{144}\)
∴ sec β = ± \(\frac{13}{12}\)
since β lies in first quadrant
∴ sec β > 0
Thus, sec β = \(\frac{13}{12}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 13

(ii)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 11

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 12

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 25.
(i) A positive acute angle is divided into two parts whose tangents are and Show that the angle is
(ii) Prove that tan 22° + tan 23° + tan 22° tan 23° = 1.
(iii) If α + β = \(\frac{\pi}{4}\), prove that (1 + tan α) (1 + tan β) = 2.
Solution:
(i) Let θ = α + β,
where tan α = \(\frac{1}{2}\)
and tan β = \(\frac{1}{3}\)
Now tan θ = tan (α + β)
= \(\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)
= \(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}\)
⇒ tan θ = 1
⇒ B = \(\frac{\pi}{4}\)
Hence the required positive acute angle be \(\frac{\pi}{4}\).

(ii) Now tan 45° = (tan 22° + tan 23°)
= \(\frac{\tan 22^{\circ}+\tan 23^{\circ}}{1-\tan 22^{\circ} \tan 23^{\circ}}\)
⇒ 1 = \(\frac{\tan 22^{\circ}+\tan 23^{\circ}}{1-\tan 22^{\circ} \tan 23^{\circ}}\)
⇒ 1 – tan 22° tan 23° = tan 22° + tan 23°
⇒ 1 = tan 22° + tan 23° + tan 22° tan 23°

(iii) Given α + β = \(\frac{\pi}{4}\)
⇒ β = \(\frac{\pi}{4}\) – α
L.H.S. = (1 + tan α) (1 + tan β)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 14

Question 26.
Prove that :
(i) tan 70° = tan 20° + 2 tan 50°
(ii) tan 56° = \(\frac{\tan 20^{\circ}+\tan 50^{\circ}}{1-\tan 20^{\circ} \tan 50^{\circ}}\)
Solution:
(i) tan 70° = tan (20° + 50°)
= \(\frac{\tan 20^{\circ}+\tan 50^{\circ}}{1-\tan 20^{\circ} \tan 50^{\circ}}\)
⇒ tan 70° [1 – tan 20° tan 50°]
= tan 20° + tan 50°
⇒ tan 70° – tan 20° tan 50° tan (90° — 70°)
= tan 20° + tan 50°
⇒ tan 70° – tan 20° tan 50° cot 20°
= tan 20° + tan 50°
⇒ tan 70° = tan 20° + 2 tan 50°

(ii) tan 56° = tan (45° + 11°)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 15

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 27.
Prove that :
(i) sin (x + y) sin (x – y) + sin (y + z) sin (y – z) + sin (z + x) sin (- x) = 0
(ii) 1 + tan x tan \(\frac{x}{2}\) = sec x = tan x cot \(\frac{x}{2}\) – 1.
Solution:
(i) L.H.S. = sin (x + y) sin (x – y) + sin (y + z) sin (y – z) + sin (z + x) sin (z – x)
= sin2x – sin2 y + sin2 y – sin2 z + sin2 z – sin2 x
= 0
= R.HS.
[∵ sin (A + B) = sin (A – B) = sin2 A – sin2 B]

(ii) L.H.S. = 1 + tan x tan \(\frac{x}{2}\)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 16

Question 28.
Prove that : tan 13x = tan 4x + tan 9x + tan 4x tan 9x tan 13x.
Solution:
Now, tan 13x = tan (4x + 9x)
= \(\frac{\tan 4 x+\tan 9 x}{1-\tan 4 x \tan 9 x}\)
⇒ tan 13x [1 – tan 4x tan 9x] = tan 4x + tan 9x
⇒ tan 13x = tan 4x + tan 9x + tan 4x tan 9x tan 13x

Question 29.
Prove that: cos 2x cos 2y + sin2 (x – y) – sin2 (x + y) = cos (2x + 2y).
Solution:
LH.S. = cos 2x cos 2y + sin2 (x – y) – sin2 (x + y)
= cos 2x cos 2y + sin (x – y + x + y) sin (x – y – x – y)
[∵ sin2 A – sin2 B = sin (A + B) sin (A – B)]
= cos 2x cos 2y + sin 2x sin (- 2y) .
= cos 2x cos 2y – sin 2x sin 2y
[∵ sin(- θ) = – sin θ]
= cos (2x + 2y)
[∵ cos A cos B – sin A sin B = cos (A + B)]
= R.H.S.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 30.
(i) cot x cot y = 2, show that \(\frac{\cos (x+y)}{\cos (x-y)}=\frac{1}{3}\).
(ii) If tan A = m tan B, prove that \(\frac{\sin (A-B)}{\sin (A+B)}=\frac{m-1}{m+1}\)
(ill) If α + β = γ and \(\frac{\tan \alpha}{\tan \beta}=\frac{x}{y}\), then prove that sin (α – β) = \(\frac{x-y}{x+y}\) sin γ.
Solution:
(i) Given cot x coty = 2
⇒ \(\frac{\cos x \cos y}{\sin x \sin y}=\frac{2}{1}\)
Applying componendo and dividendo ; we have
\(\frac{\cos x \cos y-\sin x \sin y}{\cos x \cos y+\sin x \sin y}=\frac{2-1}{2+1}\)
⇒ \(\frac{\cos (x+y)}{\cos (x-y)}=\frac{1}{3}\)

(ii) Given tan A = m tan B
⇒ \(\frac{\tan \mathrm{A}}{\tan \mathrm{B}}=\frac{m}{1}\)
⇒ \(\frac{\sin \mathrm{A} \cos \mathrm{B}}{\cos \mathrm{A} \sin \mathrm{B}}=\frac{m}{1}\)
Applying componendo and dividendo ; we have
⇒ \(\frac{\sin \mathrm{A} \cos \mathrm{B}-\cos \mathrm{A} \sin \mathrm{B}}{\sin \mathrm{A} \cos \mathrm{B}+\cos \mathrm{A} \sin \mathrm{B}}=\frac{m-1}{m+1}\)
⇒ \(\frac{\sin (\mathrm{A}-\mathrm{B})}{\sin (\mathrm{A}+\mathrm{B})}=\frac{m-1}{m+1}\) ………………(1)

(iii) Given α + β = γ ……………….(1)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 17

Question 31.
If 3 tan (θ – 15°) = tan (θ + 15°), 0 < θ < 90°, then prove that θ = 45°.
Solution:
Given, 3 tan (θ – 15°) = tan (θ + 15°)

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5 18

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.5

Question 32.
Find the range of the function 5 sin x – 12 cos x + 7.
Solution:
Let f (x) = 5 sin x – 12 cos x + 7
putting 5 = r cos α
and 12 = rsin α …………….(2)
On squaring (1) and (2) ; we have
r = \(\sqrt{25+144}\)
= \(\sqrt{169}\) = 13
On dividing (2) by (1) ; we have
tan α = \(\frac{12}{5}\)
∴ f (x) = r sin x cos α – r sin α cos x + 7
= 13 sin (x – α) + 7
since – 13 ≤ sin (x – α) ≤ 1 ∀ x ∈ R
⇒ – 13 ≤ 13 sin (x – α) ≤ 13
⇒ – 13 + 7 ≤ 13 sin (x- α) + 7 ≤ 13 + 7
⇒ – 6 ≤ f (x) ≤ 20
This range of f (x) = Rf = [- 6, 20]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4

Students often turn to ML Aggarwal Maths for Class 11 Solutions Chapter 3 Trigonometry Ex 3.4 to clarify doubts and improve problem-solving skills.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4

Question 1.
Draw the graphs of the following functions :
(i) sin 3x
(ii) 3 sin x
(iii) 2 sin 2x.
Also write their range and period.
Solution:
(i) Given y = sin 3x
= sin (3x + 2π)
= sin 3 (x + \(\frac{2 \pi}{3}\))
∴ The period of sin 3x is \(\frac{2 \pi}{3}\).
So it is suffices to draw the graph in the interval 0 to \(\frac{2 \pi}{3}\) and repeat it over other intervals.
We construct the table of values are given as under:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 1

A portion of graph is given as under
since – 1 ≤ sin 3x ≤ 1
∴ Range = {y ; y ∈ R : – 1 ≤ y ≤ 1}

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 2

(ii) Let y = 3 sin x 3 sin (x + 2π), it is defined for all x ∈ R.
Thus period of 3 sin x be 2π.
So it is sufficient to draw the graph in the interval 0 to 2π and repeat it over other intervals.
Now construct the table of values is as under :

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 3

A portion of graph is given as under :

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 4

since, – 1 ≤ sin x ≤ 1 ∀ x ∈ R
⇒ – 3 ≤ 3 sin x ≤ 3
Thus, Rf = {y ; y ∈ R, – 3 ≤ y ≤ 3}
= [- 3, 3]

(iii) Let y = 2 sin 2x. it is defined for all x ∈ R.
⇒ y = 2 sin (2x + 2π)
= 2 sin 2 (x + π)
Thus period of 2 sin 2x be π.
So it is sufficient to draw the graph in the interval 0 to π and repeat it over other intervals.
Now construct the table of values is as under:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 5

The portion of graph is represented as under :

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 6

since – 1 ≤ sin 2x ≤ 1 ∀ x ∈ R
⇒ – 2 ≤ y ≤ 2 ∀ x ∈ R
∴ range of f (x) = Rf
= {y : y ∈ R – 2 ≤ y ≤ 2} = [- 2, 2]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4

Question 2.
Draw the graph or the following functions:
(i) cos \(\frac{x}{2}\)
(ii) 3 cos 2x
(iii) 2 cos 3x
Also write their range and period.
Solution:
(i) Let y = cos \(\frac{x}{2}\)
which is defined for all x ∈ R.
Now y = cos \(\frac{x}{2}\)
= cos (\(\frac{x}{2}\) + 2π)
= cos \(\frac{1}{2}\) + (x + 4π)
∴ period of y be 4π.
So it is sufflcient to draw the graph in the interval 0 to 4π and repeat it over other intervals.
We construct the table of values is as under:
A portion of graph is given as under:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 7

since, – 1 ≤ cos \(\frac{x}{2}\) ≤ 1 ∀ x ∈ R
Thus Rf = [- 1, 1]

(ii) Let y = 3 cos 2x
= 3 cos (2x + 2π)
= 3 cos 2 (x + π), which is defined ∀ x ∈ R.
The period of 3 cos 2x be π.
So it is sufficient to draw the graph in the interval 0 to π and repeat it over other intervals.
We construct the table of values is as under:
The portion of graph is given as under:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 8

since, – 1 ≤ cos 2x ≤ 1 ∀ x ∈ R
⇒ – 3 ≤ 3 cos 2x ≤ 3
⇒ – 3 ≤ y ≤ 3
∴ Range = {y : y ∈ R ; – 3 ≤ y ≤ 3} = [- 3, 3]

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4

(iii) Let y = 2 cos 3x
= 2 cos (3x + 2π)
= 2 cos 3 (x + \(\frac{2 \pi}{3}\))
which is defined for all x ∈ R.
Thus, the period of 2 cos 3x be \(\frac{2 \pi}{3}\).
So it is sufficient to draw the graph in the interval 0 to \(\frac{2 \pi}{3}\) and then repeat it over other intervals.
We construct the table of values is given as under:

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 9

The portion of graph is given as under :

ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.4 10

since – 1 ≤ cos 3x ≤ 1 ∀ x ∈ R – 2 ≤ 2 cos 3x ≤ 2
∴ Range = [- 2, 2]