ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.3

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.3

Very short answer/objective questions (1 to 9) :

Question 1.
Write the domain of the following trigonometric functions :
(i) sin x
(ii) cos x
(ill) tan x
(iv) cot x
(v) sec x
(vi) cosec x.
Solution:
(i) f (x) = sin x is defined for all x ∈ R
∴ domain of f = R

(ii) f (x) = cos x is defined for all x ∈ R
∴ domain of f = D_f = R

(iii) f (x) = tan x
Since tan x is not defined when x be odd multiple of \(\frac{\pi}{2}\).
∴ f (x) is defined for all x ∈ R – {odd multiple of \(\frac{\pi}{2}\)}
Thus D_f = R – {(2n + 1) \(\frac{\pi}{2}\)} ; n ∈ I

(iv) Given, f (x) = cot x
Since cot x is not defined when x be an even multiple of \(\frac{\pi}{2}\) or multiple of it.
∴ D_f domain of f = R – {nπ} ∀ n ∈ I

(v) Let f (x) = sec x = \(\frac{1}{\cos x}\)
Since f (x) is not defined when cos x = 0
⇒ x = (2n + 1) \(\frac{\pi}{2}\) ∀ n ∈ I
Hence f (x) is defined for all x ∈ R except odd multiple of \(\frac{\pi}{2}\).
∴ D_f = R – {(2n + 1) \(\frac{\pi}{2}\)} ∀ n ∈ I

(vi) Let f (x) = cosec x = \(\frac{1}{\sin x}\)
since f (x) is not defined when sin x = 0
⇒ x = nπ ∀ n ∈ I
Hence f is defined for all x ∈ R except multiple of n.
∴ D_f = R – {nπ} ∀ n ∈ R.

Question 2.
Write the range of the following trigonometric functions :
(i) sin x
(ii) cos x
tan x
(iv) cot x
(v) sec x
(vi) cosec x
Solution:
(i) Since – 1 ≤ sin x ≤ 1
∴ Range of f R_f = [- 1, 1]

(ii) Let f (x) = cos x
and – 1 ≤ cos x < 1
Range of f = R_f = [- 1, 1]

(iii) Let f (x) = tan x, since tan x can take any value.
∴ Range of f = R_f = R

(iv) Let f (x) = cot x, since cot x can take any value.
∴ Range of f = R_f = R

(y) Let f (x) = sec x,
since sec x ≥ 1 or sec x ≤ – 1
∴ range of f = R_f
= (- ∞ , – 1] ∪ [1, ∞)

(vi) Let f (x) = cosec x
Since cosec x ≥ 1 or cosec x ≤ – 1
∴ Range of f = R_f
= (- ∞ , 1] ∪ [1, ∞).

Question 3.
What is the domain of the function f defined by f (x) = \(\frac{1}{3-2 \sin x}\) ?
Solution:
Given f (x) = \(\frac{1}{3-2 \sin x}\)
For D_f : f (x) must be a real number
∴ \(\frac{1}{3-2 \sin x}\) must be a real number
i.e. 3 – 2 sin x ≠ 0
⇒ sin x ≠ \(\frac{3}{2}\) which is true
since – 1 ≤ sin x ≤ 1
Thus, D_f = R.

Question 4.
Find the range of the following functions:
(i) f (x) = 2 + 5 sin 3x.
(ii) f (x) = 2 – 3 cos x
Solution:
(i) Given f (x) = 2 + 5 sin 3x
We know that – 1 ≤ sin 3x ≤ 1
⇒ – 5 ≤ sin 3x ≤ 5
⇒ 2 – 5 ≤ 2 + 5 sin 3x ≤ 2 + 5
⇒ – 3 ≤ 2 + 5 sin 3x ≤ 7
∴ range of f = R_f = [- 3, 7]

(ii) Given f (x) = 2 – 3 cos x
We know that – 1 ≤ cos x ≤ 1 ∀ x ∈ R
⇒ – 3 ≤ 3 cos x ≤ 3
⇒ 3 ≥ – 3 cos x ≥ – 3
⇒ 2 + 3 ≥ 2 – 3 cos x ≥ 2 – 3
⇒ – 1 ≤ 2 – 3 cos x ≤ 5
⇒ Range of function = R_f = [- 1, 5]

Question 5.
Which of the six trigonometric functions are positive for the angles
(i) \(\frac{4 \pi}{3}\)
(ii) – \(\frac{7 \pi}{3}\)
Solution:
(i) Since \(\frac{4 \pi}{3}\) = π + \(\frac{\pi}{3}\) clearly it lies in 3rd quadrant.
Thus tan x and cotx are positive while other t-ratios sin x, cos x, sec x and cosec x are negative.

(ii) We know that, the terminal position of 2 πn + x, n ∈ Z and x be same.
Thus – \(\frac{7 \pi}{3}\) + 2 × 2π = \(\frac{5 \pi}{3}\)
= π + \(\frac{2 \pi}{3}\)
= 2π – \(\frac{\pi}{3}\)
which lies in 4th quadrant.
Thus, cos x and sec x are positive while.
sin x, tan x, cot x and cosec x are negative.

Question 6.
In which quadrant does x lie if
(i) cos x is positive and tan x is negative
(ii) both sin x and cos x are negative
(iii) sin x = \(\frac{4}{5}\) and cos x = – \(\frac{3}{5}\)
(iv) sin x = \(\frac{2}{3}\) and cos x = – \(\frac{1}{3}\) ?
Solution:
(i) cos x is positive
∴ x lies in 1st and IVth quadrant
tan x is negative
∴ x lies in 2nd and IVth quadrant
When cosx is positive and tanx is negative.
Then x lies in 4th quadrant.

(if) Now, sin x is negative
∴ x lies in 3rd and 4th quadrant
cos x is negative ∴ x lies in 2nd and 3rd quadrant.
Now sin x and cos x both are negative
∴ x lies in III rd quadrant.

(iii) Given sin x = \(\frac{4}{5}\)
∴ x lies in I st and II nd quadrant
and cos x = – \(\frac{3}{5}\)
∴ x lies in II nd and III rd quadrant.
Now sin x = \(\frac{4}{5}\) (positive)
and cos x = – \(\frac{3}{5}\) (negative)
Then x lies in IInd quadrant.

(iv) Given sin x = \(\frac{2}{3}\) (positive)
∴ x lies in Ist and IInd quadrant
and cos x = – \(\frac{1}{3}\) (negative)
∴ x lies in 2nd and 3rd quadrant
Now Sin x = \(\frac{2}{3}\)
and cos x = – \(\frac{1}{3}\)
∴ x lies in 2nd quadrant.

Question 7.
Find the values of the following :
(i) tan \(\frac{25 \pi}{4}\)
(ii) sec \(\frac{5 \pi}{3}\)
Solution:
(i) tan \(\frac{25 \pi}{4}\) = tan (6π + \(\frac{\pi}{4}\))
= tan \(\frac{\pi}{4}\) = 1
[∵ tan (2nπ + x) = tan x ∀ n ∈ Z]

(ii) sec \(\frac{5 \pi}{3}\) = sec (2π – \(\frac{\pi}{3}\))
= sec (\(\frac{\pi}{3}\))
= \(\frac{1}{\cos \frac{\pi}{3}}\) = 2
[∵ sec (2nπ ± x) = sec x ∀ n ∈ Z]

Question 8.
Find the values of the following:
(i) cot \(\left(-\frac{7 \pi}{4}\right)\)
(ii) Sin \(\left(-\frac{17 \pi}{3}\right)\)
Solution:
(i) cot \(\left(-\frac{7 \pi}{4}\right)\) = – cot \(\left(\frac{7 \pi}{4}\right)\)
[∵ cot (- x) = – cot x]
= – cot (2π – \(\frac{\pi}{4}\))
= – cot (- \(\frac{\pi}{4}\))
[∵ cot (2π – θ) = cot (- θ)]
= + cot \(\frac{\pi}{4}\) = 1

(ii) sin \(\left(-\frac{17 \pi}{3}\right)\) = – sin \(\left(\frac{17 \pi}{3}\right)\)
[∵ cot (- θ) = – cot θ]
= – sin \(\left(4 \pi+\frac{5 \pi}{3}\right)\)
= – sin \(\left(\frac{5 \pi}{3}\right)\)
= – sin \(\left(2 \pi-\frac{\pi}{3}\right)\)
= – sin \(\left(-\frac{\pi}{3}\right)\)
[∵ sin (2π – θ) = sin (- θ)]
= sin \(\frac{\pi}{3}\)
= \(\frac{\sqrt{3}}{2}\)

Question 9.
Find the values of the following :
(i) tan 1395°
(il) cos (- 2070°).
Solution:
(i) tan 1395° = tan (1440° – 45°)
= tan (4 × 360° – 45°)
= tan (- 45°) – tan 45°
= – 1
[∵ tan (- x) = – tan x]

(ii) cos (- 2070°) = cos (2070°)
[∵ cos (- θ) = cos θ]
= cos (360° × 5 + 270°)
= cos 270°
[∵ cos (2nπ + θ) = cos θ ∀ n ∈ Z]
= cos (360° – 90°)
= cos (- 90°)
= cos 90° = 0

Short answer questions (10 to 14) :

Question 10.
If sin x = \(\frac{3}{5}\) and x lies in the second quadrant, find the value cf cos x.
Solution:
Given sin x = \(\frac{3}{5}\)
since, cos² x = 1 – sin² x
⇒ cos² x = 1 – (\(\frac{3}{5}\))²
= 1 – \(\frac{9}{25}\) = \(\frac{16}{25}\)
⇒ cos x = ± \(\frac{4}{5}\)
Since x lies in second quadrant
∴ cos x is negative.
Thus, cos x = – \(\frac{4}{5}\)

Question 11.
If cos x = – \(\frac{2}{3}\) and x lies in the third quadrant, find the value of sin x.
Solution:
Given cos x = – \(\frac{2}{3}\)
since, sin² x = 1 – cos² x
⇒ sin² x = 1 – (- \(\frac{2}{3}\))²
= 1 – \(\frac{4}{9}\) = \(\frac{5}{9}\)
⇒ sin x = ± \(\frac{\sqrt{5}}{3}\)
Since x lies in third quadrant.
∴ sin x is negative.
Thus, sin x = – \(\frac{\sqrt{5}}{3}\)

Question 12.
Find the other five trigonometric functions if
(i) cos x = – \(\frac{1}{2}\) and x lies in the third quadrant
(ii) cot x = \(\frac{3}{4}\) and x lies in the third quadrant
(iii) tan x = \(\frac{3}{4}\) and x does not lie in the first quadrant
Solution:
(i) Given cos x = – \(\frac{1}{2}\)
We know that,
sin² x + cos² x = 1
⇒ sin² x = 1 – cos² x
= 1 – (- \(\frac{1}{2}\))²
= 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∴ sin x = ± \(\frac{\sqrt{3}}{2}\)
Since x lies in third quadrant.
∴ sin x is negative.
Thus, sin x = – \(\frac{\sqrt{3}}{2}\)
∴ tan x = \(\frac{\sin x}{\cos x}\)
= \(\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\) = √3 ;
cot x = \(\frac{1}{\tan x}\)
= \(\frac{1}{\sqrt{3}}\)
sec x = \(\frac{1}{\cos x}\) = – 2
and cosec x = \(\frac{1}{\sin x}\)
= – \(\frac{2}{\sqrt{3}}\)

(ii) Given, cot x = \(\frac{3}{4}\)
We know that,
cosec² x = 1 + cot² x
∴ cosec²x = 1 + \(\left(\frac{3}{4}\right)^2\)
= 1 + \(\frac{9}{16}\)
= \(\frac{25}{16}\)
⇒ cosec x = ± \(\frac{5}{4}\)
Since x lies in third quadrant.
∴ sin x and cosec x is negative.
∴ cosec x = – \(\frac{5}{4}\)
⇒ sin x = – \(\frac{4}{5}\)
Now tan x = \(\frac{1}{\cot x}=\frac{4}{3}\)
Now cos x = cot x × sin x
= \(\frac{3}{4} \times\left(-\frac{4}{5}\right)=-\frac{3}{5}\)
Thus, sec x = – \(\frac{5}{3}\)
Hence, sin x = latex]\frac{4}{5}[/latex] ;
cos x = – \(\frac{3}{5}\);
tan x = \(\frac{4}{3}\) ;
sec x = – \(\frac{5}{3}\)
and cosec x = – \(\frac{5}{4}\)

(iii) Given tan x = \(\frac{3}{4}\) (positive)
We know that,
sec² x = 1 + tan² x
= 1 + \(\frac{9}{16}=\frac{25}{16}\)
⇒ sec x = ± \(\frac{5}{4}\) and tan x is positive.
When x lies in third quadrant
∴ cos x is negative
∴ sec x is positive
∴ sec x = – \(\frac{5}{4}\)
⇒ cos x = – \(\frac{4}{5}\)
since tan x = \(\frac{3}{4}\)
⇒ cot x = \(\frac{1}{\tan x}\)
= \(\frac{4}{3}\)
Now sin x = tan x cos x
= \(=\frac{3}{4} \times\left(-\frac{4}{5}\right)=-\frac{3}{5}\)
and cosec x = \(\frac{1}{\sin x}=-\frac{5}{3}\)

Question 13.
If sin x sec x = – 1 and x lies in the second quadrant, find sin x and sec x.
Solution:
Given sin x sec x = – 1
⇒ \(\frac{\sin x}{\cos x}\) = – 1
⇒ tan x = – 1
We know that,
sec² x = 1 + tan² x
= 1 + (- 1)²
= 1 + 1 = 2
⇒ sec x = ± √2
Since x lies in the second quadrant
∴ sec x is negative.
∴ sec x = – √2
⇒ cos x = – \(\frac{1}{\sqrt{2}}\)
Now tan x = – 1
⇒ sin x = – cos x
= – \(\left(-\frac{1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}\)

Question 14.
If tan x = – \(\frac{4}{3}\), find the value of 9 sec² x – 4 cot x.
Solution:
Given tan x = – \(\frac{4}{3}\)
We know that,
sec² x = 1 + tan² x
= 1 + \(\left(-\frac{4}{3}\right)^2\)
= 1 + \(\frac{16}{9}=\frac{25}{9}\)
∴ cot x = \(\frac{1}{\tan x}=-\frac{3}{4}\)
Thus,
9 sec² x – 4 cot x = \(9\left(\frac{25}{9}\right)-4\left(-\frac{3}{4}\right)\)
= 25 + 3 = 28

Long answer questions (15 to 23) :

Question 15.
If sec x = √2 and find the value of \(\frac{1+\tan x+\ {cosec} x}{1+\cot x-\ {cosec} x}\).
Solution:
Given sec x = √2
⇒ cos x = \(\frac{1}{\sec x}=\frac{1}{\sqrt{2}}\)
We know that,
sin² x + cos² x = 1
sin² x = 1 – cos² x
⇒ sin² x = 1 – \(\left(\frac{1}{\sqrt{2}}\right)^2\)
= \(1-\frac{1}{2}=\frac{1}{2}\)
sin x = ± \(\frac{1}{\sqrt{2}}\)
Since, \(\frac{3 \pi}{2}\) < x < 2π
x lies in fourth quadrant.
∴ sin x is negative
sin x = – \(\frac{1}{\sqrt{2}}\)
∴ cosec x = \(\frac{1}{\sin x}\) = – √2
Thus, tan x = \(\frac{\sin x}{\cos x}\)
= \(\frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\) = – 1
∴ cot x = – 1
Now, \(\frac{1+\tan x+\operatorname{cosec} x}{1+\cot x-\operatorname{cosec} x}=\frac{1-1+(-\sqrt{2})}{1-1-(-\sqrt{2})}\)
= \(-\frac{\sqrt{2}}{\sqrt{2}}\) = – 1

Question 16.
If cosec x – cot x = \(\frac{3}{2}\), find cos x. In which quadrant does x lie?
Solution:
Given cosec x – cot x = \(\frac{3}{2}\) ……………………….(1)
We know that,
cosec² x – cot² x = 1
⇒ (cosec x – cot x) (cosec x + cot x) = 1
⇒ cosec x + cot x = \(\frac{2}{3}\) [using (1)]
On adding (1) and (2) ; we have
2 cosecx = \(\frac{3}{2}+\frac{2}{3}=\frac{13}{6}\)
⇒ cosec x = \(\frac{13}{12}\)
⇒ sin x = \(\frac{12}{13}\)
eqn. (1) – eqn. (2) gives ;
– 2 cot x = \(\frac{3}{2}-\frac{2}{3}=\frac{5}{6}\)
⇒ cot x = – \(\frac{5}{12}\)
Thus, sin x is positive and cot x is negative
∴ x lies in 2nd quadrant.
Now cos x = cot x sin x
= – \(-\frac{5}{12} \times \frac{12}{13}=-\frac{5}{13}\)

Question 17.
Show that
(i) \(\sin \frac{\pi}{6} \cos 0+\sin \frac{\pi}{4} \cos \frac{\pi}{4}+\sin \frac{\pi}{3} \cos \frac{\pi}{6}=\frac{7}{4}\)
(ii) \(4 \sin \frac{\pi}{6} \sin ^2 \frac{\pi}{3}+3 \cos \frac{\pi}{3} \tan \frac{\pi}{4}+\ {cosec}^2 \frac{\pi}{2}=2 \sec ^2 \frac{\pi}{4}\)
Solution:
(i) L.H.S. = \(\sin \frac{\pi}{6} \cos 0+\sin \frac{\pi}{4} \cos \frac{\pi}{4}+\sin \frac{\pi}{3} \cos \frac{\pi}{6}\)
= \(\frac{1}{2} \times 1+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\)
= \(\frac{1}{2}+\frac{1}{2}+\frac{3}{4}\)
= \(1+\frac{3}{4}=\frac{7}{4}\)
= R.H.S.

(ii) L.H.S. = \(4 \sin \frac{\pi}{6} \sin ^2 \frac{\pi}{3}+3 \cos \frac{\pi}{3} \tan \frac{\pi}{4}+\ {cosec}^2 \frac{\pi}{2}\)
= \(4 \times \frac{1}{2} \times\left(\frac{\sqrt{3}}{2}\right)^2+3 \times \frac{1}{2} \times 1+1^2\)
= \(2 \times \frac{3}{4}+\frac{3}{2}+1\)
= \(\frac{3}{2}+\frac{3}{2}+1\) = 4
R.H.S. = 2 sec² \(\frac{\pi}{4}\)
= 2 (√2)²
= 2 × 2 = 4
Thus L.H.S. = R.H.S.

Question 18.
Taking x = \(\frac{\pi}{6}\), verify that
(i) sin² x + cos² x = 1
(ii) sin 3x = 3 sin x – 4 sin³ x
(iii) sin 2x = \(\frac{2 \tan x}{1+\tan ^2 x}\)
(iv) cos 2x = \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\)
Solution:
Given x = \(\frac{\pi}{6}\)
(i) L.H.S. = sin² x + cos² x
= sin² \(\frac{\pi}{6}\) + cos² \(\frac{\pi}{6}\)
= \(\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\)
= \(\frac{1}{4}+\frac{3}{4}\)
= 1 = R.H.S.

(ii) L.H.S. = sin 3x
= sin 3 × \(\frac{\pi}{6}\)
= sin \(\frac{\pi}{2}\) = 1
R.H.S. = 3 sin x – 4 sin³ x
= 3 sin \(\frac{\pi}{6}\) – 4 (sin \(\frac{\pi}{6}\))³
= 3 × \(\frac{1}{2}\) – 4 (\(\frac{1}{2}\))³
= \(\frac{3}{2}-\frac{1}{2}\) = 1
∴ L.H.S. = R.H.S.

(iii) L.H.S. = sin 2x
= sin 2 × \(\frac{\pi}{6}\)
= sin \(\frac{\pi}{3}\)
= \(\frac{\sqrt{3}}{2}\)
R.H.S. = \(\frac{2 \tan x}{1+\tan ^2 x}\)
= \(\frac{2 \tan \frac{\pi}{6}}{1+\tan ^2 \frac{\pi}{6}}\)
= \(-\frac{\frac{2}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)
= \(\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}\)
= \(\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}\)
= \(\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}\)
∴ L.H.S. = R.H.S.

(iv) L.H.S. = cos 2x
= cos (2 × \(\frac{\pi}{6}\))
= cos \(\frac{\pi}{3}\)
= \(\frac{1}{2}\)
and R.H.S. = \(\frac{1-\tan ^2 x}{1+\tan ^2 x}\)
= \(\frac{1-\left(\tan \frac{\pi}{6}\right)^2}{1+\left(\tan \frac{\pi}{6}\right)^2}\)
= \(\frac{1-\left(\frac{1}{\sqrt{3}}\right)^2}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)
= \(\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{4}{3}}=\frac{1}{2}\)
∴ L.H.S. = R.H.S.

Question 19.
Taking x = \(\frac{\pi}{3}\) and y = \(\frac{\pi}{6}\), verify that
(i) sin (x + y) ≠ sin x + sin y
(ii) cos (x + y) ≠ cos x + cos y.
Solution:
Given x = \(\frac{\pi}{3}\)
and y = \(\frac{\pi}{6}\)
(i) L.H.S. = sin (x + y)
= sin \(\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\)
= sin \(\frac{\pi}{2}\) = 1
R.H.S. = sin x + sin y
= sin \(\frac{\pi}{3}\) + sin \(\frac{\pi}{6}\)
= \(\frac{\sqrt{3}}{2}+\frac{1}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
L.H.S. ≠ R.H.S.
Thus, sin (x + y) ≠ sin x + sin y

(ii) L.H.S. = cos (x + y)
= cos \(\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\)
= cos \(\frac{\pi}{2}\) = 0
and R.H.S. = cos x + cos y
= cos \(\frac{\pi}{3}\) +cos \(\frac{\pi}{6}\)
= \(\frac{1}{2}+\frac{\sqrt{3}}{2}\)
= \(\frac{\sqrt{3}+1}{2}\)
∴ L.H.S. ≠ R.H.S.
Thus, cos (x + y) ≠ cos x + cos y.

Question 20.
Find the value of x (0 < x < \(\frac{\pi}{2}\)) satisfying
(i) \(\frac{\cos x}{\ {cosec} x+1}+\frac{\cos x}{\ {cosec} x-1}\) = 2
(ii) \(\frac{\cos ^2 x-3 \cos x+2}{\sin ^2 x}\) = 1
(iii) 2 sin² x = \(\frac{1}{2}\)
(iv) 3 cos x = 2 sin² x
(v) tan² x – (1 + √3) tan x + √3 = 0. (ISC 2020)
Solution:
(i) Given,

(ii) \(\frac{\cos ^2 x-3 \cos x+2}{\sin ^2 x}\) = 1
⇒ cos² x – 3 cos x + 2 = sin² x
= 1 – cos² x
⇒ 2 cos² x – 3 cos x + 1 = 0
cos x = \(\frac{3 \pm \sqrt{9-8}}{4}\)
= \(\frac{3 \pm 1}{4}\)
= 1, \(\frac{1}{2}\)
either cos x = 1 or cos x = \(\frac{1}{2}\)
⇒ x = 0 or x = \(\frac{\pi}{3}\)
But 0 < x < \(\frac{\pi}{2}\)
∴ x = \(\frac{\pi}{3}\)

(iii) Given 2 sin² x =
⇒ sin² x = \(\frac{1}{4}\)
⇒ sin x = ± \(\frac{1}{2}\)
But 0 < x < \(\frac{\pi}{2}\)
∴ x lies in first quadrant.
Thus, sin x is positive
⇒ sin x = \(\frac{1}{2}\)
= sin \(\frac{\pi}{6}\)
⇒ x = \(\frac{\pi}{6}\) ∈ (0, \(\frac{\pi}{2}\))

(iv) Given 3 cos x = 2 sin² x
⇒ 3 cos x = 2 (1 – cos² x)
⇒ 2 cos² x + 3 cos x – 2 = 0
⇒ cos x = – 2, \(\frac{1}{2}\)
But – 1 ≤ cos x ≤ 1 as 0 < x < \(\frac{\pi}{2}\)
∴ cos x is positive.
Thus cos x = \(\frac{1}{2}\)
⇒ x = \(\frac{\pi}{3}\)

(v) Given trigonometrical eqn. be
tan² x – (1 + √3) tan x + √3 = 0
⇒ tan x (tan x – 1) – √3 (tan x – 1) = 0
⇒ (tan x – √3) (tan x – 1) = 0
⇒ tan x =√3 0r tan x = 1
⇒ x = \(\frac{\pi}{3}\) or x = \(\frac{\pi}{4}\)
[∵ 0 < x < \(\frac{\pi}{2}\)]
⇒ x = \(\frac{\pi}{4}\), \(\frac{\pi}{3}\)

Question 21.
If x, y ∈R, show that the equation 2xy sin² θ = x² + y² is possible only when x = y ≠ 0.
Solution:
Given, 2xy sin² θ = x² + y²
sin² θ = \(\frac{x^2+y^2}{2 x y}\) ………………….(1)
We know that,
sin² θ ≥ 0 and x² + y² ≥ 0
Thus, eqn. (1) holds if xy > 0 …………………..(2)
Also, sin² θ ≤ 1
⇒ \(\frac{x^2+y^2}{2 x y}\) ≤ 1
⇒ x² + y² ≤ 2xy [∵ xy > 0]
(x – y)² ≤ 0 but x, y ∈ R
Also from (2) ; xy > 0
∴ none of x, y can be zero.
Thus the given equation is possible only when x = y ≠ 0.

Question 22.
Find the least values of the following functions :
(i) sin² x + cosec² x
(ii) tan² x + cot² x.
Solution:
(i) sin² x + cosec² x
= (sin x – cosec x)² + 2 sin x cosec x
= (sin x – cosec x)² + 2 ≥ 2
[ (sin x – cosec x)² ≥ 0 ∀ x ∈ R]
∴ least value of sin² x + cosec² x = 2

(ii) tan² x + cot² x = (tan x – cot x)² + 2 tan x cot x
= (tan x – cot x)² + 2 ≥ 2
[: (tan x -cot x)² ≥ 0 ∀ x ∈ R]
Thus, least value of tan 2x + cot 2x is equal to 2.

Question 23.
Find the domain and the range of the following functions :
(i) 5 – 4 sin 3x
(ii) \(\frac{1}{4-\cos 3 x}\)
(iii) |cos 2x|
(iv) [1 + 2 sin 3x]
Solution:
(i) Let f (x) = 5 – 4 sin 3x
For domain of f (x) ; f (x) must be a real number
⇒ 5 – 4 sin 3x must be a real number
which is clearly a real number ∀ x ∈ R
∴ D_f = R
We know that,
– 1 ≤ sin 3x ≤ 1
⇒ 4 ≥ – 4 sin 3x ≥ – 4
⇒ 5 + 4 ≥ 5 – 4 sin 3x ≥ 5 – 4
⇒ 1 ≤ 5 – 4 sin 3x ≤ 9
∴ range of f (x) = [1, 9]

(ii) Let f(x) = \(\frac{1}{4-\cos 3 x}\)
since, – 1 ≤ cos 3x ≤ 1 ∀ x ∈ R
= 1 ≥ – cos 3x ≥ – 1
= 4 + 1 ≥ 4 – cos 3x ≥ 4 – 1
3 ≤ 4 — cos 3x ≤ 5
4 cos 3x > 0 ∀ x ∈ R
For D_f : f (x) must be a real number,
which is only possible if 4 – cos 3x ≠ 0
which is true for all ∀ x ∈ R
[∵ 4 – cos 3x > 0]
Thus D_f = R
since, – 1 ≤ cos 3x ≤ 1
⇒ 1 ≥ – cos 3x ≥ – 1
⇒ 4 + 1 ≥ 4 – cos 3x ≥ 4 – 1
⇒ 3 ≤ 4 – cos 3x ≤ 5
⇒ \(\frac{1}{3} \geq \frac{1}{4-\cos 3 x} \geq \frac{1}{5}\)
⇒ \(\frac{1}{5}\) ≤ f (x) ≤ \(\frac{1}{3}\)
Thus, range of f (x) = [\(\frac{1}{5}\), \(\frac{1}{3}\)]

(iii) Let f (x) = [cos 2x]
For D_f : f (x) must be a real number
⇒ [cos 2x] must be a real number
⇒ cos 2x must be a real number,
which is a real number ∀ x ∈ R
Thus D_f = R
For R_f :
since – 1 ≤ cos 2x ≤ 1 ∀ x ∈ R
⇒ [cos 2x] = – 1, 0, 1
∴ R_f = {- 1, 0, 1}

(iv) Let f (x) = [1 + 2 sin 3x]
For D_f : f (x) must be a real number
⇒ [1 + 2 sin 3x] must be a real number
⇒ 1 + 2 sin 3x must be a real number, which is a real number ∀ x ∈ R.
Thus, D_f = R
For R_f :
We know that,
⇒ – 1 ≤ sin 3x ≤ 1 x ∈ R
⇒ – 2 ≤ 2 sin 3x ≤ 2
⇒ 1 – 2 ≤ 1 + 2 sin 3x ≤ 1 + 2
⇒ – 1 ≤ 1 + 2 sin 3x ≤ 3
∴ [1 + 2 sin 3x] = – 1, 0, 1, 2, 3
Thus, R_f = {- 1, 0, 1, 2, 3}