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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 = Df = 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 Df = 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.
∴ Df 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 $$\frac{\pi}{2}$$.
∴ Df = 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.
∴ Df = 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 Rf = [- 1, 1]

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

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

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

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

(vi) Let f (x) = cosec x
Since cosec x ≥ 1 or cosec x ≤ – 1
∴ Range of f = Rf
= (- ∞ , 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 Df : 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, Df = 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 = Rf = [- 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 = Rf = [- 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 – $$\frac{\pi}{3}$$
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
∴ 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
∴ Df = 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 Df : 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 Df = 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 Df : 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 Df = R
For Rf :
since – 1 ≤ cos 2x ≤ 1 ∀ x ∈ R
⇒ [cos 2x] = – 1, 0, 1
∴ Rf = {- 1, 0, 1}

(iv) Let f (x) = [1 + 2 sin 3x]
For Df : 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, Df = R
For Rf :
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, Rf = {- 1, 0, 1, 2, 3}