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

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)

(ii)

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.

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.

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

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

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°

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°

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)

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

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

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

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

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:

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 \(\frac{x}{2}\) = – \(\sqrt{\frac{1+\cos x}{2}}\)
= – \(\sqrt{\frac{1+\left(-\frac{4}{5}\right)}{2}}\)
= \(-\sqrt{\frac{1}{10}}=-\frac{1}{\sqrt{10}}\)

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)

(ii)

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

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

(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 = \(\frac{1}{\sqrt{2}}\)

⇒ 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)

(iv)

(v)

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

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.