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

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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
∴ D_f = R
and g(x) = cos^-1 2x ;
– 1 ≤ 2x ≤ 1
– \(\frac{1}{2}\) ≤ x ≤ \(\frac{1}{2}\)
∴ D_g = [- \(\frac{1}{2}\), \(\frac{1}{2}\)]
Thus dom (f(x) + g(x)) = D_f ∩ D_g
= 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 = \(\frac{2}{\sqrt{5}}\)