Regular engagement with Understanding ISC Mathematics Class 12 Solutions Chapter 2 Inverse Trigonometric Functions MCQs can boost students confidence in the subject.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 2 Inverse Trigonometric Functions MCQs

Choose the correct answer from the given four options in questions (1 to 35) :

Question 1.
The domain of the function sin-1 $$\sqrt{x-1}$$ is
(a) [- 1, 1]
(b) [1, 2]
(c) [0, 1]
(d) none of these
Solution:
(b) [1, 2]

We know that, domain of sin-1 $$\sqrt{x-1}$$ ;
– 1 ≤ $$\sqrt{x-1}$$ ≤ 1 but $$\sqrt{x-1}$$ ≥ 0
⇒ 0 ≤ x – 1 ≤ 1
⇒ 1 ≤ x ≤ 2
⇒ x ∈ [1, 2].

Question 2.
the domain of the function cos-1 (- x2) is
(a) (0, 1)
(b) [0, 1]
(c) [- 1, 1]
(d) Φ
Solution:
(c) [- 1, 1]

For domain of cos-1 (- x2) ;
– 1 ≤ – x2 ≤ 1 but – x2 ≤ 0
⇒ – 1 ≤ – x2 ≤ 0
⇒ 1 ≥ x2 ≥ 0
⇒ |x| ≤ 1
⇒ – 1 ≤ x ≤ 1
⇒ x ∈ [- 1, 1]

Question 3.
The domain of the function sin-1 (x2 – 4) is
(a) [3, 5]
(b [- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$]
(c) [- √5, – √3] ∩ [√3, √5]
(d) [- √5, – √3] ∪ [√3, √5]
Solution:
(d) [- √5, – √3] ∪ [√3, √5]

For domain of sin-1 (x2 – 4) ;
– 1 ≤ x2 – 4 ≤ 1
⇒ 4 – 1 ≤ x2 ≤ 1 + 4
⇒ 3 ≤ x2 ≤ 5
⇒ √3 ≤ |x| ≤ √5
⇒ – √5 ≤ x ≤ – √3
⇒ √3 ≤ x ≤ √5
⇒ x ∈ [- √5, – √3] ∪ [√3, √5].

Question 4.
The domain of the function sin 2x + cos-1 2x is
(a) [- $$\frac{1}{2}$$, $$\frac{1}{2}$$]
(b) [- 1, 1]
(c) R
(d) [- 1, π + 1]
Solution:
(a) [- $$\frac{1}{2}$$, $$\frac{1}{2}$$]

Let f(x) = sin 2x
∴ Df = R
and g(x) = cos-1 2x ;
– 1 ≤ 2x ≤ 1
– $$\frac{1}{2}$$ ≤ x ≤ $$\frac{1}{2}$$
∴ Dg = [- $$\frac{1}{2}$$, $$\frac{1}{2}$$]
Thus dom (f(x) + g(x)) = Df ∩ Dg
= R ∩ [- $$\frac{1}{2}$$, $$\frac{1}{2}$$]
= [- $$\frac{1}{2}$$, $$\frac{1}{2}$$]

Question 5.
The Range of the principal value of the branch sec-1 x is
(a) (- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$)
(b) [- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$] – [0, π]
(c) [0, π] – {$$\frac{\pi}{2}$$}
(d) (0, π)
Solution:
(c) [0, π] – {$$\frac{\pi}{2}$$} [by def.]

Question 6.
The simplified value of sin ($$\frac{\pi}{2}$$ – sin-1 (- $$\frac{\sqrt{3}}{2}$$)) is
(a) $$\frac{1}{2}$$
(b) $$\frac{1}{\sqrt{2}}$$
(c) $$\frac{\sqrt{3}}{2}$$
(d) – $$\frac{\sqrt{3}}{2}$$
Solution:
(a) $$\frac{1}{2}$$

sin-1 (- $$\frac{\sqrt{3}}{2}$$)) = x,
– $$\frac{\pi}{2}$$ ≤ x ≤ $$\frac{\pi}{2}$$
⇒ sin x = – $$\frac{\sqrt{3}}{2}$$
= – sin $$\frac{\pi}{3}$$
= sin (- $$\frac{\pi}{3}$$)
⇒ x = – $$\frac{\pi}{3}$$ ∈ [- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$]
∴ sin ($$\frac{\pi}{2}$$ – sin-1 (- $$\frac{\sqrt{3}}{2}$$)) = sin ($$\frac{\pi}{2}$$ – (- $$\frac{\pi}{3}$$))
= sin ($$\frac{\pi}{2}$$ + $$\frac{\pi}{3}$$)
= cos $$\frac{\pi}{3}$$
= $$\frac{1}{2}$$.

Question 7.
The principal value of sin-1 (sin (- $$\frac{10 \pi}{3}$$)) is
(a) $$\frac{\pi}{3}$$
(b) – $$\frac{\pi}{3}$$
(c) $$\frac{2 \pi}{3}$$
(d) – $$\frac{2 \pi}{3}$$
Solution:
(a) $$\frac{\pi}{3}$$

∵ sin-1 (sin θ) = θ if θ ∈ [- $$\frac{\pi}{2}$$,$$\frac{\pi}{2}$$]
∴ sin-1 (sin (- $$\frac{10 \pi}{3}$$)) ≠ $$\frac{10 \pi}{3}$$
Thus, sin-1 (sin (- $$\frac{10 \pi}{3}$$)) = sin-1 {- sin ($$\frac{10 \pi}{3}$$)}
= – sin-1 {sin (2π + $$\frac{4 \pi}{3}$$)}
[∵ sin-1 (- x) = – sin-1 x]
= – sin-1 {sin $$\frac{4 \pi}{3}$$}
= – sin-1 {sin (π + $$\frac{\pi}{3}$$)}
= – sin-1 {- sin $$\frac{\pi}{3}$$}
= sin-1 (sin $$\frac{\pi}{3}$$)
= $$\frac{\pi}{3}$$

Question 8.
The principal value of cos-1 (cos (- $$\frac{34 \pi}{9}$$)) is
(a) $$\frac{34 \pi}{9}$$
(b) $$\frac{\pi}{9}$$
(c) $$\frac{2 \pi}{9}$$
(d) $$-\frac{2 \pi}{9}$$
Solution:
(c) $$\frac{2 \pi}{9}$$

cos-1 (cos (- $$\frac{34 \pi}{9}$$)) = cos-1 {cos $$\frac{34 \pi}{9}$$}
= cos-1 {cos (4π – $$\frac{2 \pi}{9}$$)}
= cos-1 {cos (- $$\frac{2 \pi}{9}$$)}
= cos-1 (cos $$\frac{2 \pi}{9}$$)
= $$\frac{2 \pi}{9}$$ ∈ [0, π]

Question 9.
The principal value of sin-1 (cos ($$\frac{43 \pi}{9}$$)) is
(a) $$\frac{2 \pi}{5}$$
(b) – $$\frac{2 \pi}{5}$$
(c) $$\frac{\pi}{10}$$
(d) – $$\frac{\pi}{10}$$
Solution:
(d) – $$\frac{\pi}{10}$$

sin-1 (cos ($$\frac{43 \pi}{9}$$)) = sin-1 {cos (8π + $$\frac{43 \pi}{5}$$)}
= sin-1 {cos $$\frac{3 \pi}{5}$$}
= sin-1 {sin ($$\frac{\pi}{2}$$ – $$\frac{3 \pi}{5}$$)}
= $$\frac{\pi}{2}$$ – $$\frac{3 \pi}{5}$$
= $$\frac{\pi}{10}$$

Question 10.
The principal value of tan-1 (tan (- 6)) is
(a) – 6
(b) 2π – 6
(c) 6 – 2π
(d) none of these
Solution:
(b) 2π – 6

Clearly – 6 ∉ (- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$)
But 2π – 6 ∈ (- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$)
∴ tan-1 {tan (- 6)} = tan-1 {tan (2π – 6)}
[∵ tan (2π – θ) = tan (- θ)]
= 2π – 6.

Question 11.
sin-1 (cos x) = $$\frac{\pi}{2}$$ – x is valid for
(a) – π ≤ x ≤ 0
(b) 0 ≤ x ≤ π
(c) – $$\frac{\pi}{2}$$ ≤ x ≤ $$\frac{\pi}{2}$$
(d) – $$\frac{\pi}{4}$$ ≤ x ≤ $$\frac{3 \pi}{4}$$
Solution:
(b) 0 ≤ x ≤ π

sin-1 (cos x) = sin-1 {sin ($$\frac{\pi}{2}$$ – x)}
= $$\frac{\pi}{2}$$ – x
[∵ sin-1 (sin θ) = θ ∀ θ ∈ [- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$]]
i.e., when – $$\frac{\pi}{2}$$ ≤ $$\frac{\pi}{2}$$ – x ≤ $$\frac{\pi}{2}$$
⇒ – π ≤ – x ≤ 0
⇒ π ≥ x ≥ 0
⇒ 0 ≤ x ≤ π
∴ x ∈ [0, π]

Question 12.
cos-1 (sin x) = $$\frac{\pi}{2}$$ – x is valid for
(a) – π ≤ x ≤ 0
(b) 0 ≤ x ≤ π
(c) – $$\frac{\pi}{2}$$ ≤ x ≤ $$\frac{\pi}{2}$$
(d) none of these
Solution:
(c) – $$\frac{\pi}{2}$$ ≤ x ≤ $$\frac{\pi}{2}$$

cos-1 (sin x) = cos-1 {cos ($$\frac{\pi}{2}$$ – x)}
= $$\frac{\pi}{2}$$ – x
if 0 ≤ $$\frac{\pi}{2}$$ – x ≤ π
if – $$\frac{\pi}{2}$$ ≤ – x ≤ $$\frac{\pi}{2}$$
if – $$\frac{\pi}{2}$$ ≤ x ≤ $$\frac{\pi}{2}$$
if x ∈ [- $$\frac{\pi}{2}$$ , $$\frac{\pi}{2}$$]

Question 13.
If sin-1 x + sin-1 y = $$\frac{2 \pi}{3}$$, then cos-1 x + cos-1 y is equal to
(a) π
(b) $$\frac{2 \pi}{3}$$
(c) $$\frac{\pi}{3}$$
(d) $$\frac{\pi}{6}$$
Solution:
(c) $$\frac{\pi}{3}$$

We know that,
sin-1 x + cos-1 y = $$\frac{\pi}{2}$$ ∀ |x| ≤ 1
Given sin-1 x + sin-1 y = $$\frac{2 \pi}{3}$$
⇒ $$\frac{\pi}{2}$$ – cos-1 x + $$\frac{\pi}{2}$$ – cos-1 y = $$\frac{2 \pi}{3}$$
⇒ cos-1 x + cos-1 y = π – $$\frac{2 \pi}{3}$$
= $$\frac{\pi}{3}$$

Question 14.
If ∝ ≤ tan-1 x + cot-1 x + sin-1 x ≤ β, then
(a) ∝ = 0, β = π
(b) ∝ = 0, β = $$\frac{\pi}{2}$$
(c) ∝ = $$\frac{\pi}{2}$$, β = π
(d) ∝ = – $$\frac{\pi}{2}$$, β = $$\frac{\pi}{2}$$
Solution:
(a) ∝ = 0, β = π

Given, ∝ ≤ tan-1 x + cot-1 x + sin-1 x ≤ β
⇒ ∝ ≤ $$\frac{\pi}{2}$$ + sin-1 x ≤ β
[∵ tan-1 x + cot-1 x = $$\frac{\pi}{2}$$ ∀ x ∈ R]
⇒ ∝ – $$\frac{\pi}{2}$$ ≤ sin-1 x ≤ β – $$\frac{\pi}{2}$$
Also, – $$\frac{\pi}{2}$$ ≤ sin-1 x ≤ $$\frac{\pi}{2}$$
∴ ∝ – $$\frac{\pi}{2}$$ = – $$\frac{\pi}{2}$$
⇒ ∝ = 0
and β – $$\frac{\pi}{2}$$ = $$\frac{\pi}{2}$$
⇒ β = π

Question 15.
The value tan (cos-1 $$\frac{3}{5}$$ + tan-1 $$\frac{1}{4}$$) is
(a) $$\frac{3}{4}$$
(b) $$\frac{19}{12}$$
(c) $$\frac{8}{19}$$
(d) $$\frac{19}{8}$$
Solution:
cos-1 $$\frac{3}{5}$$ = tan-1 $$\frac{4}{3}$$

∴ tan (cos-1 $$\frac{3}{5}$$ + tan-1 $$\frac{1}{4}$$) = tan (tan-1 $$\frac{4}{3}$$ + tan-1 $$\frac{1}{4}$$)
= tan {tan-1 $$\left(\frac{\frac{4}{3}+\frac{1}{4}}{1-\frac{4}{3} \times \frac{1}{4}}\right)$$}
[∵ tan-1x – tan-1 y = tan-1 $$\left(\frac{x-y}{1+x y}\right)$$ if xy < 1]
= tan {tan-1 $$\left(\frac{\frac{19}{12}}{\frac{8}{12}}\right)$$}
= tan {tan-1 $$\frac{19}{8}$$}
= $$\frac{19}{8}$$
[∵ tan (tan-1 x) = x ∀ x ∈ R]

Question 16.
The value of cot (cosec-1 $$\frac{5}{3}$$ + tan-1 $$\frac{2}{3}$$) is
(a) $$\frac{6}{17}$$
(b) $$\frac{3}{17}$$
(c) $$\frac{4}{17}$$
(d) $$\frac{5}{17}$$
Solution:
(a) $$\frac{6}{17}$$

∴ cot (cosec-1 $$\frac{5}{3}$$ + tan-1 $$\frac{2}{3}$$) = cot (tan-1 $$\frac{3}{4}$$ + tan-1 $$\frac{2}{3}$$)

= cot {tan-1 $$\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \times \frac{2}{3}}\right)$$}
= cot {tan-1 $$\left(\frac{\frac{17}{12}}{\frac{6}{12}}\right)$$}
= cot {tan-1 $$\frac{17}{6}$$}
= cot {cot-1 $$\frac{6}{17}$$}
[∵ cot-1 x = tan-1 $$\frac{1}{x}$$ ; x > 0]
= $$\frac{6}{17}$$

Question 17.
The value of tan-1 5 + tan-1 3 – cot-1 $$\frac{4}{7}$$ is
(a) – $$\frac{\pi}{2}$$
(b) $$\frac{\pi}{2}$$
(c) 0
(d) π
Solution:
(b) $$\frac{\pi}{2}$$

tan-1 5 + tan-1 3 = tan-1 $$\left(\frac{5+3}{1-5 \times 3}\right)$$ + π
[∵ tan-1 x + tan-1 y = π + tan-1 $$\left(\frac{x+y}{1-x y}\right)$$ if xy > 1, x, y > 0]
= π + tan-1 (- $$\frac{8}{14}$$)
∴ tan-1 5 + tan-1 3 – cot-1 $$\frac{4}{7}$$ = π + tan-1 (- $$\frac{8}{14}$$) – cot-1 $$\frac{4}{7}$$
= π – tan-1 $$\frac{8}{14}$$ – tan-1 $$\frac{7}{4}$$
[∵ cot-1 x = tan-1 $$\frac{1}{x}$$ ; x > 0
and tan-1 (- x) = – tan-1 x]
= π – {tan-1 ($$\frac{4}{7}$$) + tan-1 $$\frac{7}{5}$$}
= π – {tan-1 $$\frac{4}{7}$$ + cot-1 $$\frac{4}{7}$$}
= π – $$\frac{\pi}{2}$$
= $$\frac{\pi}{2}$$
[∵ tan-1 x + cot-1 x = $$\frac{\pi}{2}$$ ∀ x ∈ R]

Question 18.
The value of tan (cos-1 (- $$\frac{2}{7}$$) – $$\frac{\pi}{2}$$) is
(a) $$\frac{2}{3 \sqrt{5}}$$
(b) $$\frac{2}{3}$$
(c) $$\frac{1}{\sqrt{5}}$$
(d) $$\frac{4}{\sqrt{5}}$$
Solution:
(a) $$\frac{2}{3 \sqrt{5}}$$

tan (cos-1 (- $$\frac{2}{7}$$) – $$\frac{\pi}{2}$$) = tan {π – cos-1 $$\frac{2}{7}$$ – $$\frac{\pi}{2}$$) = tan {π – cos-1 $$\frac{2}{7}$$ – $$\frac{\pi}{2}$$}
[∵ cos-1 (- x) = π – cos-1 x]
= tan ($$\frac{\pi}{2}$$ – cos-1 $$\frac{2}{7}$$)
Let cos-1 $$\frac{2}{7}$$ = θ
⇒ cos θ = $$\frac{2}{7}$$

= tan ($$\frac{\pi}{2}$$ – θ)
= cot θ = $$\frac{2}{\sqrt{45}}=\frac{2}{3 \sqrt{5}}$$

Question 19.
The value of cot (sin-1 x) is
(a) $$\frac{1}{x}$$
(b) $$\frac{\sqrt{1+x^2}}{x}$$
(c) $$\frac{x}{\sqrt{1+x^2}}$$
(d) $$\frac{\sqrt{1-x^2}}{x}$$
Solution:
put sin-1 x = θ
⇒ x = sin θ
cot (sin-1 x) = cot θ
= $$\frac{\sqrt{1-x^2}}{x}$$

Question 20.
The value of cot (cos-1 ($$\frac{7}{25}$$)) is
(a) $$\frac{25}{24}$$
(b) $$\frac{24}{25}$$
(c) $$\frac{7}{24}$$
(d) $$\frac{25}{7}$$
Solution:
(c) $$\frac{7}{24}$$

Let cos-1 $$\frac{7}{25}$$ = θ
⇒ cos θ = $$\frac{7}{25}$$
∴ cot {cos-1 $$\frac{7}{25}$$} = cot θ
= $$\frac{\cos \theta}{\sin \theta}$$
= $$\frac{\frac{7}{25}}{\sqrt{1-\left(\frac{7}{25}\right)^2}}$$
= $$\frac{\frac{7}{25}}{\frac{24}{25}}=\frac{7}{24}$$

Question 21.
The value of tan (2 sin-1 $$\frac{1}{\sqrt{3}}$$) is
(a) 2√2
(b) 2√3
(c) √2
(d) $$\frac{2}{\sqrt{3}}$$
Solution:
(a) 2√2

Let sin-1 $$\frac{1}{\sqrt{3}}$$ = θ
⇒ sin θ = $$\frac{1}{\sqrt{3}}$$
∴ tan θ = $$\frac{1}{\sqrt{2}}$$

∴ tan (2 sin-1 $$\frac{1}{\sqrt{3}}$$) = tan 2θ
= $$\frac{2 \tan \theta}{1-\tan ^2 \theta}$$
= $$\frac{2 \times \frac{1}{\sqrt{2}}}{1-\left(\frac{1}{\sqrt{2}}\right)^2}$$
= $$\frac{\sqrt{2}}{1-\frac{1}{2}}$$
= 2√2

Question 22.
The value of tan ($$\frac{1}{2}$$ cos-1 $$\frac{2}{\sqrt{5}}$$) is
(a) 2 + √5
(b) √5 – 2
(c) 5 + √2
(d) 2 – √5
Solution:
(b) √5 – 2

Let cos-1 $$\frac{2}{\sqrt{5}}$$ = θ
⇒ cos θ = $$\frac{2}{\sqrt{5}}$$
∴ tan ($$\frac{1}{2}$$ cos-1 $$\frac{2}{\sqrt{5}}$$) = tan $$\frac{\theta}{2}$$
= $$\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}$$
= $$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$$
= $$\sqrt{\frac{1-\frac{2}{\sqrt{5}}}{1+\frac{2}{\sqrt{5}}}}$$
= $$\sqrt{\frac{\sqrt{5}-2}{\sqrt{5}+2}}$$
= $$\sqrt{\frac{(\sqrt{5}-2)^2}{5-4}}$$
= √5 – 2

Question 23.
The value of tan2 (sec-1 2 + cot2 (cosec-1 3) is
(a) 5
(b) 11
(c) 13
(d) 15
Solution:
(b) 11

sec-1 2 = tan-1 $$\left(\frac{\sqrt{3}}{1}\right)$$

cosec-1 3 = cot-1 $$\left(\frac{\sqrt{8}}{1}\right)$$

∴ tan2 (sec-1 2) + cot2 (cosec-1 3) = {tan (sec-1 2)}2 + {cot (cosec-1 3)}2
= {tan (tan-1 √3)2 + {cot (cot-1 √8)2
= (√3)2 + (√8)2
= 3 + 8 = 11

Question 24.
The value of x which satisfies the equation tan-1 x = sin-1 $$\left(\frac{3}{\sqrt{10}}\right)$$ is
(a) – $$\frac{1}{3}$$
(b) $$\frac{1}{3}$$
(c) – 3
(d) 3
Solution:
(d) 3

Given tan-1 x = sin-1 $$\left(\frac{3}{\sqrt{10}}\right)$$
put sin-1 $$\frac{3}{\sqrt{10}}$$ = θ
⇒ sin θ = $$\frac{3}{\sqrt{10}}$$

∴ tan-1 x = θ
⇒ x = tan θ
⇒ x = 3.

Question 25.
If 4 sin-1 x + cos-1 x = π, then the value of x is
(a) $$\frac{2}{3}$$
(b) $$\frac{1}{3}$$
(c) $$\frac{1}{2}$$
(d) 2
Solution:
(c) $$\frac{1}{2}$$

Given 4 sin-1 x + cos-1 x = π
⇒ 3 sin-1 x + sin-1 x + cos-1 x = π
⇒ 3 sin-1 x + $$\frac{\pi}{2}$$ = π
[∵ sin-1 x + cos-1 x = $$\frac{\pi}{2}$$ ∀ x ∈[- 1, 1]]
⇒ sin-1 x = $$\frac{\pi}{6}$$
⇒ x = sin $$\frac{\pi}{2}$$
= $$\frac{1}{2}$$

Question 26.
If cos (sin-1 $$\frac{2}{5}$$ + cos-1 x) = 0, then x is equal to
(a) 0
(b) $$\frac{1}{5}$$
(c) $$\frac{2}{5}$$
(d) 1
Solution:
(c) $$\frac{2}{5}$$

Given cos {sin-1 $$\frac{2}{5}$$ + cos-1 x} = 0
⇒ sin-1 $$\frac{2}{5}$$ + cos-1 x = $$\frac{\pi}{2}$$
⇒ sin-1 $$\frac{2}{5}$$ = $$\frac{\pi}{2}$$ – cos-1 x = sin-1 x
⇒ $$\frac{2}{5}$$ = x

Question 27.
If sin-1 $$\frac{x}{5}$$ + cosec-1 $$\frac{5}{4}$$ = $$\frac{\pi}{2}$$, then x is equal to
(a) 1
(b) 3
(c) 4
(d) 5
Solution:
(b) 3

Given sin-1 $$\frac{x}{5}$$ + cosec-1 $$\frac{5}{4}$$ = $$\frac{\pi}{2}$$
⇒ sin-1 $$\frac{x}{5}$$ = $$\frac{\pi}{2}$$ – cosec-1 $$\frac{5}{4}$$
= sec-1 $$\frac{5}{4}$$
[∵ sec-1 x + cosec-1 x = $$\frac{\pi}{2}$$, |x| ≥ 1]
⇒ sin-1 $$\frac{x}{5}$$ = sin-1 $$\left(\frac{3}{5}\right)$$
⇒ $$\frac{x}{5}=\frac{3}{5}$$
⇒ x = 3

Question 28.
The simplified value of sin (cot-1 (cos (tan-1 1))) is
(a) 0
(b) 1
(c) $$\frac{1}{\sqrt{3}}$$
(d) $$\sqrt{\frac{2}{3}}$$
Solution:
(d) $$\sqrt{\frac{2}{3}}$$

sin (cot-1 (cos (tan-1 1))) = sin (cot-1 (cos $$\frac{\pi}{4}$$))
[∵ tan-1 1 = tan-1 (tan $$\frac{\pi}{4}$$) = $$\frac{\pi}{4}$$]

= sin (cot-1 $$\frac{1}{\sqrt{2}}$$)
put cot-1 $$\frac{1}{\sqrt{2}}$$ = θ
⇒ cot θ = $$\frac{1}{\sqrt{2}}$$
= sin θ
= $$\frac{\sqrt{2}}{\sqrt{3}}$$

Question 29.
The equation 2 cos-1 x + sin-1 x = $$\frac{11 \pi}{6}$$ has
(a) no solution
(b) unique solution
(c) two solutions
(d) more than two solutions
Solution:
(a) no solution

Given 2 cos-1 x + sin-1 x = $$\frac{11 \pi}{6}$$ …………..(1)
⇒ cos-1 x + (cos-1 x + sin-1 x) = $$\frac{11 \pi}{6}$$
⇒ cos-1 x + $$\frac{\pi}{2}$$ = $$\frac{11 \pi}{6}$$
[∵ sin-1 x + cos-1 x = $$\frac{\pi}{2}$$ , |x| ≤ 1]
⇒ cos-1 x = $$\frac{11 \pi}{6}-\frac{\pi}{2}$$
= $$\frac{8 \pi}{6}=\frac{4 \pi}{3}$$
x = cos $$\frac{4 \pi}{3}$$
= cos (π + $$\frac{\pi}{3}$$)
= – cos $$\frac{\pi}{3}$$
⇒ x = – $$\frac{1}{2}$$
But x = – $$\frac{1}{2}$$ does not satisfies the eqn. (1)
Since 2 cos-1 x + sin-1 x = 2 cos-1 (- $$\frac{1}{2}$$) + sin-1 (- $$\frac{1}{2}$$)
= 2 [π – cos-1 $$\frac{1}{2}$$] – sin-1 $$\frac{1}{2}$$
= 2 [π – $$\frac{\pi}{3}$$] – $$\frac{\pi}{6}$$
= $$\frac{4 \pi}{3}$$ – $$\frac{\pi}{6}$$
= $$\frac{7 \pi}{6}$$ ≠ R.H.S.
∴ given eqn. has no solution.

Question 30.
If cos-1 x + cos-1 y + cos-1 z = 3π, then the value of x2 + y2 + z2 – 2xyz is
(a) 1
(b) 3
(c) 6
(d) 5
Solution:
(d) 5

We know that, 0 ≤ cos-1 x ≤ π
so maximum value of cos-1 x = π
∴ cos-1 x + cos-1 y + cos-1 z = 3π
⇒ cos-1 x = π; cos-1 y = π; cos-1 z = π
⇒ x = cos π = – 1 ;
y = cos π = – 1
and z = cos π = – 1
∴ x2 + y2 + z2 – 2xyz = (- 1)2 + (- 1)2 + (- 1)2 – 2 × (- 1) (- 1) (- 1)
= 1 + 1 + 1 + 2 = 5

Question 31.
Which of the following is true?
(a) Domain of sin1 x is [- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$]
(b) Range of cos (sin-1 x + cos-1 x) is {- 1, 1}
(c) Range of sin (sin-1 x + cos-1 x) = {1}
(d) Range of cos-1 x is (0, π)
Solution:
(c) Range of sin (sin-1 x + cos x) = {1}

since Domain of sin-1 x ; x ∈ [- 1, 1]
cos (cos’ x + sin x) = cos $$\frac{\pi}{2}$$ = 0
∴ Range of cos (cos-1 x + sin-1 x) be {0}
∴ Range of sin (sin-1 x + cos-1 x)
= sin $$\frac{\pi}{2}$$ = 1 be {1}.
and Range of cos-1 x be [0, π]

Question 32.
If sin-1 $$\left(\frac{2 a}{1+a^2}\right)$$ + cos-1 $$\left(\frac{1-a^2}{1+a^2}\right)$$ = tan-1 $$\left(\frac{2 x}{1-x^2}\right)$$, where a, x ∈ (0, 1), then the value of x is
(a) 0
(b) a
(c) $$\frac{a}{2}$$
(d) $$\left(\frac{2 a}{1-a^2}\right)$$
Solution:
(d) $$\left(\frac{2 a}{1-a^2}\right)$$

We know that,
2 tan-1 x = sin-1 $$\left(\frac{2 x}{1x^2}\right)$$
= cos-1 $$\left(\frac{1-x^2}{1+x^2}\right)$$
Given sin-1 $$\left(\frac{2 a}{1+a^2}\right)$$ + cos-1 $$\left(\frac{1-a^2}{1+a^2}\right)$$ = tan-1 $$\left(\frac{2 x}{1-x^2}\right)$$
⇒ 2 tan-1 a + 2 tan-1 a = 2 tan-1 x
⇒ 4 tan-1 a = 2 tan-1 x
⇒ tan-1 x = 2 tan-1 a
= tan-1 $$\left(\frac{2 a}{1-a^2}\right)$$
⇒ x = $$\frac{2 a}{1-a^2}$$

Question 33.
The principal value of tan-1 (tan $$\frac{3 \pi}{5}$$) is
(a) $$\frac{2 \pi}{5}$$
(b) $$-\frac{2 \pi}{5}$$
(c) $$\frac{3 \pi}{5}$$
(d) $$-\frac{3 \pi}{5}$$
Solution:
(b) $$-\frac{2 \pi}{5}$$

We know that,
tan-1 (tan θ) = θ
θ ∈ (- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$)
∴ tan-1 (tan $$\frac{3 \pi}{5}$$) ≠ $$\frac{3 \pi}{5}$$ as $$\frac{3 \pi}{5}$$ ∉ (- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$)
∴ tan-1 (tan $$\frac{3 \pi}{5}$$) = tan-1 (tan (π – $$\frac{2 \pi}{5}$$))
= tan-1 (- tan $$\frac{2 \pi}{5}$$)
= – tan-1 (tan $$\frac{2 \pi}{5}$$)
[∵ tan-1 (- x) = – tan-1 x]
= – $$\frac{2 \pi}{5}$$
[∵ tan-1 (tan θ) = θ ∀ θ ∈(- $$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$)]

Question 34.
tan-1 3 + tan-1 λ = tan-1 $$\left(\frac{3+\lambda}{1-3 \lambda}\right)$$ is valid for what value of λ?
(a) λ ∈ (- $$\frac{1}{3}$$, $$\frac{1}{3}$$)
(b) λ > $$\frac{1}{3}$$
(c) λ < $$\frac{1}{3}$$
(d) all real values of λ
Solution:
(c) λ < $$\frac{1}{3}$$

We know that,
tan-1 x + tan-1 y = tan-1 $$\left(\frac{x+y}{1-x y}\right)$$ if xy < 1
Given tan-1 3 + tan-1 λ = tan-1 $$\left(\frac{3+\lambda}{1-3 \lambda}\right)$$
if 3λ < 1 if λ < $$\frac{1}{3}$$

Question 35.
If cos (sin-1 $$\frac{2}{\sqrt{5}}$$ + cos-1 x) = 0, then x is equal to
(a) $$\frac{1}{\sqrt{5}}$$
(b) – $$\frac{2}{\sqrt{5}}$$
(c) $$\frac{2}{\sqrt{5}}$$
(d) 1
Solution:
(c) $$\frac{2}{\sqrt{5}}$$

Given cos (sin-1 $$\frac{2}{\sqrt{5}}$$ + cos-1 x) = 0
⇒ sin-1 + cos-1 x =
⇒ sin-1 $$\frac{2}{\sqrt{5}}$$ = $$\frac{\pi}{2}$$ – cos-1 x = sin-1 x
⇒ x = $$\frac{2}{\sqrt{5}}$$