OP Malhotra Class 12 Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4

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S Chand Class 12 ICSE Maths Solutions Chapter 4 Inverse Trigonometric Functions Ex 4

Question 1.
Write down the values of :
(i) sin^-1\(\frac{\sqrt{3}}{2}\)
(ii) cos^-1(\(\frac { 1 }{ 2 }\))
(iii) tan^-1 1
(iv) tan^-1 0
(v) cot^-1\(\left(\frac{2}{\sqrt{3}}\right)\)
(vi) Sec^-1\(\frac { 1 }{ 2 }\)
(vii) cosec^-1 2
(viii) cos^-1(-\(\frac { 1 }{ 2 }\))
Solution:

Question 2.
Find :
(i) cos A, if cos^-1\(\frac { 1 }{ 2 }\) = A
(ii) cosec A, if sin^-1\(\frac { 1 }{ 3 }\) = A
(iii) sin A, if tan^-1(\(\frac { 1 }{ 3 }\)) = A
(iv) θ, if tan^-1 3 = θ
(v) cot θ, if tan^-1\(\sqrt{3}\) = θ
(vi) x, if sin^-1(\(\frac { 1 }{ 2 }\)) = tan^-1x
Solution:

Question 3.
Find the principal value of each of the following:
(i) sin (sin^-1 \(\frac { 1 }{ 2 }\))
(ii) tan^-1 tan(\(\frac { π }{ 6 }\))
(iii) cot^-1(tan^-1\(\frac { 4 }{ 5 }\))
(iv) sin^-1(cos(\(\frac { π }{ 4 }\))
(v) sin(cos^-1 \(\frac { 1 }{ 2 }\))
(vi) cos (cot^-1(-\(\sqrt{3}\)))
(vii) sin(2 sin^-1\(\frac { 2 }{ 3 }\))
(viii) cos^-1 (sin 220°)
(ix) sin(\(\frac { 1 }{ 2 }\) cos^-1\(\frac { 4 }{ 5 }\))
(x) tan [sin^-1 (-1)]
(xi) tan^-1(cot\(\frac { 4π }{ 3 }\))
(xii) sin(tan^-1 1) + cos(cos^-1\(\frac { 1 }{ 2 }\))
(xiii) tan\(\left(\sin ^{-1} \frac{\sqrt{2}}{2}\right)-\cot \left(\cos ^{-1} \frac{\sqrt{2}}{2}\right)\)
(xiv) tan^-1(\(\left(-\frac{\sqrt{3}}{3}\right)\))
(xv) cosec^-1(-\(\left(-\frac{2 \sqrt{3}}{3}\right)\)))
(xvi) cos^-1[sin (tan^-1 (-1)]
(xvii) sin(2 tan^-1 3)
Solution:
(i) sin (sin^-1 \(\frac { 1 }{ 2 }\))
[∵ sin(sin^-1x) = x ∀ x ∈ [-1, 1]]

(ii) tan^-1 tan(\(\frac { π }{ 6 }\)) = \(\frac { π }{ 6 }\)
[∵ tan^-1(tan x) = x ∀ x ∈ [0, π]]
Here, \(\frac { π }{ 6 }\) ∈(\(\frac { -π }{ 2 }\), \(\frac { π }{ 2 }\))

(iii) First of all, we convert tan^-1 to cot^-1. For this, we construct a right triangle with p = 4 & b = 5
∴ h = \(\sqrt{b^2+p^2}\)
= \(\sqrt{s^2+4^2}=\sqrt{41}\)
Thus, \(\tan ^{-1} \frac{4}{5}=\cot ^{-1} \frac{5}{4}\)
∴ \(\cot \left(\tan ^{-1} \frac{4}{5}\right)=\cot \left(\cot ^{-1} \frac{5}{4}\right)=\frac{5}{4}\)
[∵ cot (cot^-1x) = x ∀ x ∈ R]

(iv) sin^-1\(\left(\cos \frac{\pi}{4}\right)\)

(v) sin(cos^-1 \(\frac { 1 }{ 2 }\)) = sin\(\left(\cos ^{-1} \cos \frac{\pi}{3}\right)\)
= \(\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\)
[∵ cos^-1(cos x) = x ∀ x ∈ [- 0, π]

(vi) Let cot^-1(-\(\sqrt{3}\)) = y ; y ∈ (0, π)

(vii) Let sin^-1\(\frac { 2 }{ 3 }\) = θ ⇒ \(\frac { 2 }{ 3 }\) = sin θ
∴ cos θ = \(\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}\)
∵ θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
Thus sin(2sin^-1\(\frac { 2 }{ 3 }\)) = sin 2θ
= 2 sin θ cos θ
= 2 x \(\frac{2}{3} \times \frac{\sqrt{5}}{3}=\frac{4 \sqrt{5}}{9}\)

(viii) cos^-1(sin 220°)
= cos^-1{sin(180° + 40°}
= cos^-1{- sin 40°}
= π – cos^-1 {sin 40°}
[∵ cos^-1(- x) = π – cos^-1 x ∀ x ∈ [- 1, 1]]
= 180° – cos^-1{cos(\(\frac { π }{ 2 }\) – 40°)}
= 180° – (90° – 40°) = 130°
[∵ cos^-1(cos x) = x ∀ x ∈ [0, π]]

(ix) Put cos^-1\(\frac { 4 }{ 5 }\) = θ ⇒ cos θ = \(\frac { 4 }{ 5 }\)
where θ ∈ [0, π]

(x) tan[sin^-1(-1)]= tan[-sin^-1 1]
[∵ sin^-1(-x) = – sin^-1x ∀x ∈ [-1, 1]]
= tan\(\left[-\frac{\pi}{2}\right]=-\tan \frac{\pi}{2} \rightarrow-\infty\)

(xi) \(\tan ^{-1}\left(\cot \frac{4 \pi}{3}\right)=\tan ^{-1}\left\{\cot \left(\pi+\frac{\pi}{3}\right)\right\}\)

(xii) sin(tan^-1 1) + cos(cos^-1 \(\frac { 1 }{ 2 }\))
= \(\sin \frac{\pi}{4}+\frac{1}{2}\)
[∵ cos^-1(cos^-1) = x ∀ x ∈ [- 1, 1]]
= \(\frac{1}{\sqrt{2}}+\frac{1}{2}=\frac{2+\sqrt{2}}{2 \sqrt{2}}=\frac{\sqrt{2}+1}{2}\)

(xiii) \(\tan \left(\sin ^{-1} \frac{\sqrt{2}}{2}\right)-\cot \left(\cos ^{-1} \frac{\sqrt{2}}{2}\right)\)

(xiv) tan^-1(\(\left(-\frac{\sqrt{3}}{3}\right)\))

Question 4.
Verify the following :
(i) \(\sin ^{-1} \frac{\sqrt{2}}{2}-\sin ^{-1} \frac{1}{2}=\frac{\pi}{12}\)
(ii) \(\cos ^{-1} 0+\tan ^{-1}(-1)=\tan ^{-1} 1\)
Solution:

Question 5.
Show that
(i) 2 sin^-1 x = sin^-1(2x\(\sqrt{1-x^2}\))
(ii) 2 cos^-1 x = cos^-1 (2x² – 1) if 0 ≤ x ≤ 1
(iii) 3sin^-1 = sin^-1 (3x – 4x³) if – \(\frac { 1 }{ 2 }\) ≤ x \(\frac { 1 }{ 2 }\)
(iv) 3cos^-1 x = cos^-1(4x³ – 3x) if \(\frac { 1 }{ 2 }\) ≤ x ≤ 1
(v) sin^-1(- x) = – sin^-1 x
(vi) cos^-1 (- x) = π – cos^-1 x
(vi) tan^-1(- x) = – tan^-1 x
Solution:
(i) Put sin^-1x = θ
⇒ x = sin θ; θ ∈[ \(\frac { -π }{ 2 }\), \(\frac { π }{ 2 }\) ]
∴ R.H.S = sin^-1(2x\(\sqrt{1-x^2}\))
= sin^-1(2 sin θ \(\sqrt{1-\sin ^2 \theta}\) )
= sin^-1(2 sin θ |cos θ|)
= sin^-1(2 sin θ cos θ)
[∵ cos θ > 0 ∀ θ ∈ [ \(\frac { -π }{ 2 }\), \(\frac { π }{ 2 }\) ];
Thus |cos θ| = cos θ]
= sin^-1 (sin 2θ)
= 2θ = 2 sin^-1x = L.H.S.

(ii) Put cos^-1x = θ ⇒ x = cos θ
Now 0 ≤ x ≤ 1 ⇒ 0 ≤ cos θ ≤ 1
⇒ 0 ≤ θ ≤ \(\frac { π }{ 2 }\)
∴ R.H.S. = cos^-1(2x² – 1)
= cos^-1(2 cos²θ – 1)
= cos^-1(cos 2θ) = 2θ
= 2 cos^-1 x = L.H.S.
[∵0 ≤ θ ≤ \(\frac { π }{ 2 }\) ⇒ 0 ≤ 2θ ≤ π ⇒ 2θ ∈ [0, π]
Thus cos^-1(cos 2θ) = 2θ ∀θ ∈ [0, π]]

(iii) Let sin^-1 x = θ ⇒ x = sin θ

(iv) Put cos^-1x = θ ⇒ x = cos θ
since \(\frac { 1 }{ 2 }\) ≤ x ≤ 1 \(\frac { 1 }{ 2 }\) ≤ cos θ ≤ 1
⇒ 0 ≤ θ ≤ \(\frac { π }{ 3 }\)
R.H.S.= cos^-1(4x³ – 3x)
= cos^-1 {4 cos³ θ – 3 cos θ}
= cos^-1 {cos 3θ} = 3θ
= 3 cos^-1x = L.H.S.
[∵0 ≤ θ ≤ \(\frac { π }{ 3 }\) ⇒ 0 ≤ 3θ ≤ π,
Thus cos^-1(cos x) = x ∀ x ∈ [0, π]]

(v) Let sin^-1(- x) = θ ⇒ – x = sin θ
⇒ x = – sin θ = sin (- θ)
⇒ sin^-1 x = – θ ⇒ θ = – sin^-1 x
⇒ sin^-1 (-x) = – sin^-1x

(vi) Let cos^-1(-x) = θ ⇒ – x = cos θ
⇒ x = – cos θ = cos (π – θ)
⇒ cos^-1 x = π – θ
⇒ π – cos^-1(-x)
⇒ cos^-1 (-x) = π – cos^-1x

(vii) Let tan^-1(- x) = θ ⇒ – x = tan θ
⇒ x = – tan θ = + tan(- θ)
⇒ tan^-1(x) = – θ
⇒ tan^-1 x = – tan^-1(- x)
⇒ tan^-1(- x) = – tan^-1 x

Question 6.
Show that :
(i) sin\(\left(\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{4}{5}\right)=\frac{63}{65}\)
(ii) sin\(\left(\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}\right)=1\)
(iii) tan\(\left(\sin ^{-1} \frac{\sqrt{3}}{2}-\cos ^{-1} \frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{3}\)
(iv) cos\(\left(\tan ^{-1} \frac{15}{8}-\sin ^{-1} \frac{7}{25}\right)=\frac{297}{425}\)
(v) sin\(\left(\sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{3}{2}\right)=\frac{3+4 \sqrt{3}}{10}\)
(vi) 2 tan^-1\(\frac { 1 }{ 2 }\) = tan^-1\(\frac { 4 }{ 3 }\)
Solution:
(i) sin^-1 \(\frac { 5 }{ 13 }\) + sin^-1 \(\frac { 4 }{ 5 }\)

(ii) sin\(\left(\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}\right)\) = sin \(\frac { π }{ 2 }\) = 1
[∵ tan^-1(x) + cot^-1x = \(\frac { π }{ 2 }\) ∀ x ∈ R]

(iii) tan\(\left(\sin ^{-1} \frac{\sqrt{3}}{2}-\cos ^{-1} \frac{\sqrt{3}}{2}\right)\)
= \(\tan \left[\frac{\pi}{3}-\frac{\pi}{6}\right]=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}\)

(iv) First of all, we convert tan^-1 to cos^-1. For this we construct a right triangle with b = 8, p = 15

(v) First of all, we convert cos^-1 to sin^-1, for this we construct a right triangle with b = 3 and h = 5
∴ p = \(\sqrt{h^2-b^2}=\sqrt{25-9}=4\)
Thus, \(\cos ^{-1} \frac{3}{5}=\sin ^{-1}\left(\frac{4}{5}\right)\)
∴ \(\quad \sin ^{-1} \frac{1}{2}+\cos ^{-1} \frac{3}{5}=\sin ^{-1} \frac{1}{2}+\sin ^{-1} \frac{4}{5}\)
= \(\sin ^{-1}\left[\frac{1}{2} \sqrt{1-\frac{16}{25}}+\frac{4}{5} \sqrt{1-\frac{1}{4}}\right]\)

Question 7.
Simplify:
(i) sin (2 cos^-1 x)
(ii) cos (2 sin^-1 x)
(iii) tan (sin^-1 y)
(iv) y = \(\frac { 1 }{ 2 }\)tan^-1(x + π)
Solution:
(i) Put cos^-1 x = θ
⇒ x = cos θ & θ ∈ [0, π]
∴ sin θ = \(\sqrt{1-\cos ^2 \theta}=\sqrt{1-x^2}\)
[∵ sin θ > 0 ∀ x ∈ [0, π]]
Thus, sin (2 cos^-1x) = sin(2θ)
= 2 sin θ cos θ
= 2x \(\sqrt{1 – x²}\)

(ii) Put sin^-1x = θ ⇒ sin θ = x;
where θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
∴ cos θ > 0
& cos θ = \(\sqrt{1-\sin ^2 \theta}=\sqrt{1-x^2}\)
∴ cos(2sin^-1x) = cos 2θ = 2 cos0178θ – 1
= 2\(\left(\sqrt{1-x^2}\right)^2\)
= 2(1 – x²) – 1
= 1 – 2x²

Question 8.
Solve the following for x in terms of y:
(i) y = 2sin^-13x
(ii) y = 3cos^-12x
(iii) y = \(\frac { 1 }{ 2 }\)tan^-1(x + π)
Solution:
(i) Given y = 2 sin^-1 3x …(1)
Let sin^-13x = θ ⇒ 3x = sin θ
∴ from(1); y = 2θ ⇒ θ = \(\frac { y }{ 2 }\)
⇒ sin θ = sin \(\frac { y }{ 2 }\) ⇒ 3x = sin \(\frac { y }{ 2 }\)
⇒ x = \(\frac { 1 }{ 3 }\) sin \(\frac { y }{ 2 }\)

(ii) Given y = 3 cos^-1 2x ⇒ \(\frac { y }{ 3 }\) = cos^-12x
⇒ 2x = cos\(\frac { y }{ 3 }\) ⇒ x = \(\frac { 1 }{ 2 }\) cos\(\frac { y }{ 3 }\)

(iii) Given y = \(\frac { 1 }{ 2 }\) tan^-1 (x + π)
⇒ 2y = tan^-1(x + π)
⇒ tan 2y = tan {tan^-1(x + π)} = x + π
[∵ tan (tan^-1x) = x ∀ x ∈ R]
⇒ x = tan 2y – π

Question 9.
Prove that :
(i) \(\cos ^{-1} \frac{4}{5}=\tan ^{-1} \frac{3}{4}\)
(ii) \(\tan ^{-1} 2-\tan ^{-1} 1=\tan ^{-1} \frac{1}{3}\)
(iii) \(2 \tan ^{-1} \frac{1}{3}=\tan ^{-1} \frac{3}{4}\)
(iv) \(2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}\)
(v) \(\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\)
(vi) \(\sin ^{-1} \frac{4}{5}+2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{2}\)
(vii) \(\cos ^{-1} \frac{4}{5}+\cot ^{-1} \frac{5}{3}=\tan ^{-1} \frac{27}{11}\)
(viii) \(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\cos ^{-1} \frac{2}{\sqrt{5}}\)
(ix) \(\dot{2}\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)=\tan ^{-1} \frac{4}{3}\)
Solution:
(i) We convert cos^-1 to tan^-1 for this we construct a right triangle with b = 4 & h = 5
∴ P = \(\sqrt{h^2-b^2}=\sqrt{25-16}\) = 3
Thus, cos^-1\(\frac { 4 }{ 5 }\) = tan^-1\(\frac { 3 }{ 4 }\)

[we convert cos^-1 to sin^-1 for this we construct a right triangle with b = 1 and h = 3

[We convert tan^-1 to cos^-1, for this we construct a right triangle with p = 1; b = 2

Question 10.
Prove that
\(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\)
Solution:

Prove the following :

Question 11.
4(cos^-13 + cosec^-1\(\sqrt{5}\)) = π
Solution:

Question 12.
\(\cos ^{-1} \frac{63}{65}+2 \tan ^{-1} \frac{1}{5}=\sin ^{-1} \frac{3}{5}\)
Solution:

Now we convert cos^-1 to tan^-1 for this, we construct a right triangle with b = 63; h = 65

Question 13.
\(\tan ^{-1}\left(\frac{1}{2} \tan 2 . A\right)+\tan ^{-1}(\cot \mathrm{A})\) + tan^-1 (cot A³ A) = 0
Solution:

Question 14.
If tan^-1x + tan^-1y + tan^-1z = \(\frac { π }{ 2 }\). show that xy + yz + zx = 1.
Solution:

Solve for x for the following equations:

Question 15.
cos^-1 x + sin^-1\(\frac { x }{ 2 }\) = \(\frac { π }{ 6 }\) (Dhanbad)
Solution:

[∵ cos(cos^-1x) = x ∀ x ∈ [-1, 1] & sin(cos^-1x) = \(\sqrt{1-x^2}\) ]
⇒ \(\sqrt{1-x^2}\) = 0 ⇒ x² = 1
⇒ x = ± 1
But x = – 1 does not satisfies eqn. (1).
where x = – 1;
L.H.S. = cos^-1(- 1) + sin^-1(-\(\frac { 1 }{ 2 }\))
= π + (-\(\frac { π }{ 6 }\) = \(\frac { 5π }{ 6 }\) & RHS. = \(\frac { π }{ 6 }\)
⇒ L.H.S. ≠ R.H.S.
Thus, x = 1 be the only solution.

Question 16.
sin^-1 x + sin^-1 2x = \(\frac { π }{ 3 }\)
Solution:
Given eqn. be,
sin^-1 x + sin^-1 2x = \(\frac { π }{ 3 }\) … (1)
⇒ sin^-12x = \(\frac { π }{ 3 }\) – sin^-1x
⇒ 2x = sin(\(\frac { π }{ 3 }\) – sin^-1x)
⇒ 2x = sin\(\frac { π }{ 3 }\)cos(sin^-1x) – cos\(\frac { π }{ 3 }\)sin(sin^-1x)
⇒ 2x = \(\frac{\sqrt{3}}{2} \sqrt{1-x^2}-\frac{1}{2} x\)
[∵ cos(sin^-1x) = \(\sqrt{1-x^2}\) & sin(sin^-1x) = x ∀ x ∈ [-1, 1]]
⇒ \(\frac{5 x}{2}=\frac{\sqrt{3}}{2} \sqrt{1-x^2}\); on squaring both sides
⇒ 25x² = 3(1 – x²) ⇒ 28x² = 3
⇒ x = ± \(\sqrt{\frac{3}{28}}= \pm \frac{1}{2} \sqrt{\frac{3}{7}}\)
When x = – \(\frac{1}{2} \sqrt{\frac{3}{7}}\), L.H.S. of eqn. (1) is negative while R.H.S. is positive.
Thus, the only solution be x = \(\frac{1}{2} \sqrt{\frac{3}{7}}\).

Question 17.
tan^-1 2x + tan^-13x = \(\frac { π }{ 4 }\)
Solution:
Given eqn. be, tan^-1 2x + tan^-13x = \(\frac { π }{ 4 }\)

Hence x = \(\frac { 1 }{ 6 }\) be the only solution.

Question 18.
sin\(\left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)\) = 1.
Solution:

Question 19.
sin\(\left(\frac{1}{5} \cos ^{-1} x\right)\) = 1.
Solution:
sin\(\left(\frac{1}{5} \cos ^{-1} x\right)\) = 1.
Let \(\frac { 1 }{ 2 }\) cos^-1 x = y ⇒ cos^-1x = 5y
since 0 ≤ cos^-1 x ≤ π ⇒ 0 ≤ 5y ≤ π
⇒ 0 ≤ y ≤ \(\frac { π }{ 5 }\)
∴ sin y ≠ 1
Thus, the given equations has no solution.

Question 20.
\(\tan ^{-1} \frac{1}{2 x+1}+\tan ^{-1}-\frac{1}{4 x+1}=\tan ^{-1} \frac{2}{x^2}\)
Solution:
Given,

Question 21.
\(\tan ^{-1}\left(\frac{x-2}{x-4}\right)+\tan ^{-1}\left(\frac{x+2}{x+4}\right)=\frac{\pi}{4}\)
Solution:

Question 22.
2 tan^-1(cosx) = tan^-1(2 cosec x)
Solution:
Given eqn. be,

Question 23.
Evaluate the following
(i) sin cot^-1 cos tan^-1 x
(ii) tan\(\left[\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right]\)
(iii) \(\cos ^{-1} x+\cos ^{-1}\left[\frac{x}{2}+\frac{\sqrt{3-3 x^2}}{2}\right], \frac{1}{2} \leq x \leq 1\).
Solution:
(i) Let tan^-1x = y ⇒ x = tan y
Now see y = \(\sqrt{1+\tan ^2 y}=\sqrt{1+x^2}\)

(iii) Let cos^-1 x = θ ⇒ x = cos θ

Question 24.
Solve for x : cos(sin^-1 x) = \(\frac { 1 }{ 9 }\).
Solution:
Given cos(sin^-1x) = \(\frac { 1 }{ 9 }\)
put sin^-1 x = θ ⇒ x = sin θ
& θ ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
∴ cos θ > 0
Thus cos θ = \(\frac { 1 }{ 9 }\) ⇒ \(\sqrt{1-\sin ^2 \theta}=\frac{1}{9}\)
⇒ \(\sqrt{1-x^2}=\frac{1}{9}\)
on squaring both sides, we have
1 – x² = \(\frac { 1 }{ 81 }\) ⇒ x² = 1 – \(\frac { 1 }{ 81 }\) = \(\frac { 80 }{ 81 }\)
⇒ x = ± \(\sqrt{\frac{80}{81}}= \pm \frac{4 \sqrt{5}}{9}\)

Question 25.
Find the value of
\(\tan \frac{1}{2}\left[\sin ^{-1} \frac{2 x}{1+x^2}+\cos ^{-1} \frac{1-y^2}{1+y^2}\right]\),
|x| < 1, y > 0 and xy < 1.
Solution:
Now, put x = tan θ ⇒ θ = tan^-1x
∴\(\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)\)
= sin^-1(sin 2θ) = 2θ = 2 tan^-1x
[since – 1 < x < 1 ⇒ – 1 < tan θ < 1

Question 26.
Solve the equation
\(\sin ^{-1} 6 x+\sin ^{-1}(6 \sqrt{3} x)=-\frac{\pi}{2}\)
Solution:
Given eqn. be,

on squaring ; we have
⇒ 36x² = 1 – 108 x²
⇒ 144 x² = 1 ⇒ x = ± \(\frac { 1 }{ 12 }\)
Clearly eqn. (1) only satisfied when x < 0
Thus, x = – \(\frac { 1 }{ 12 }\)

Question 27.
If sin[cot^-1(x + 1)] = cos(tan^-1 x), then find x.
Solution:
Given,
sin[cot^-1(x + 1)] = cos(tan^-1x) … (1)
we convert cot^-1 to sin^-1, for this we construct a right triangle with b = x + 1 ; P = 1

Now we convert tan^-1 to cos^-1, we construct a right triangle with p = x, b = 1
∴ h = \(\sqrt{b^2+h^2}=\sqrt{1+x^2}\)
∴ tan^-1x = cos^-1 \(\frac{1}{\sqrt{1+x^2}}\) … (3)
using eqn. (2) & eqn (3) in eqn (1); we have
\(\sin \left[\sin ^{-1}\left(\frac{1}{\sqrt{x^2+2 x+2}}\right)\right]=\cos \left[\cos ^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right)\right]\)
⇒ \(\frac{1}{\sqrt{x^2+2 x+2}}=\frac{1}{\sqrt{1+x^2}}\) [∵ sin(sin^-1 x) = x & cos(cos^-1x) = x ∀ x ∈ [-1,1] ]
on squaring both sides; we have
1 + x² = x² + 2x + 2 ⇒ 2x = – 1 ⇒ x = – 1/2

Question 28.
Show that \(\tan ^{-1}\left(\frac{1}{\sqrt{3}} \tan \frac{x}{2}\right)=\frac{1}{2} \cos ^{-1}\left(\frac{1+2 \cos x}{2+\cos x}\right)\)
Solution:

Question 29.
If cos^-1\(\frac { x }{ 2 }\) + cos^-1\(\frac { y }{ 3 }\) = θ, prove that 9x² – 12xycosθ + 4y² = 36sin²θ.
Solution:
Given \(\cos \frac{x}{2}+\cos ^{-1} \frac{y}{2}\) = θ

On squaring both sides; we have
⇒ \(\left(\frac{x y}{6}-\cos \theta\right)^2=\left(1-\frac{x^2}{4}\right)\left(1-\frac{y^2}{9}\right)\)
⇒ (xy – 6cos θ)² = (4 – x²)(9 – y²)
⇒ x²y² + 36 cos² θ – 12xy cos θ = 36 – 4y² – 9x² + x²y²
⇒ 9x² – 12xy cos θ + 4 y² = 36(1 – cos² θ) = 36 sin² θ

Question 30.
If cos^-1 x + cos^-1 y + cos^-1 z = π, prove that x² + y² + z² + 2xyz = 1.
Solution:
Given eqn. be,
cos^-1 x + cos^-1y + cos^-1z = π
⇒ cos^-1 x + cos^-1y = π – cos^-1z
⇒ cos^-1 \(\left\{x y-\sqrt{1-x^2} \sqrt{1-y^2}\right\}\) = π – cos^-1z
⇒ xy – \(\sqrt{1-x^2} \sqrt{1-y^2}\)
= cos(π – cos^-1z)
⇒ xy – \(\sqrt{1-x^2} \sqrt{1-y^2}\) = – cos(cos^-1z)
⇒ xy + z = \(\sqrt{1-x^2} \sqrt{1-y^2}\) = – z
[cos(cos^-1 x) = x ∀ |x| ≤ 1]
on squraing both sides, we have
⇒ (xy + z)² = (1 – x²)(1 – y²)
⇒ x²y² + z² + 2xyz = 1 – x² – y² + x²y²
⇒ x² + y² + z² + 2xyz = 1,
which is the required result.

Examples

Question 1.
Prove that
\(\sin ^{-1} \frac{x}{\sqrt{1+x^2}}+\cos ^{-1} \frac{x+1}{\sqrt{x^2+2 x+2}}\) = tan^-1(x² + x + 1)

Solution:
Let \(\sin ^{-1} \frac{x}{\sqrt{1+x^2}}=\theta\)
⇒ sin θ = \(\frac{x}{\sqrt{1+x^2}}\)
we construct a right triangle with p = x
& h = \(\sqrt{1+x^2}\)
∴ b = \(\sqrt{h^2-p^2}=\sqrt{1+x^2-x^2}\) = 1
∴ sin^-1 \(\bar { E }\) = tan^-1x
we convert cos^-1 to tan^-1, for this we construct a right triangle with b = x + 1

Question 2.
Solve the equation
\(\tan ^{-1}(x+2)+\tan ^{-1}(2-x)=\tan ^{-1}\left(\frac{2}{3}\right)\)
Solution:

Question 3.
Show that \(\sin ^{-1} \frac{\sqrt{3}}{2}+2 \tan ^{-1} \frac{1}{\sqrt{3}}=\frac{2 \pi}{3}\)
Solution:
L.H.S
= \(\sin ^{-1} \frac{\sqrt{3}}{2}+2 \tan ^{-1} \frac{1}{\sqrt{3}}\)
= \(\sin ^{-1}\left(\sin \frac{\pi}{2}\right)+2 \tan ^{-1}\left\{\tan \frac{\pi}{6}\right\}\)
= \(\frac{\pi}{3}+2 \times \frac{\pi}{6}=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2 \pi}{3}\)
[∵ \(\sin ^{-1}(\sin x)=x \forall x \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \& \tan ^{-1}(\tan x)=x \forall x \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\) ]

Question 4.
\(\sin ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\cot ^{-1}\left(\frac{1}{11}\right)\)
Solution:
Firstly we convert sin-1 to tan-1, for this, we construct a right triangle with p = 4 & h = 5
∴ \(\sqrt{h^2-p^2}=\sqrt{25-16}\) = 3

Thus, \(\sin ^{-1}\left(\frac{4}{5}\right)=\tan ^{-1}\left(\frac{4}{3}\right)\)
Now we convert cos-1 to tan-1, for this we construct a right triangle with b – 2 & h = \(\sqrt{5}\)
∴ P = \(\sqrt{h^2-b^2}=\sqrt{5-4}\) = 1
Thus, \(\cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)=\tan ^{-1}\left(\frac{1}{2}\right)\)

Question 5.
\(2\left(\tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}\right)\) = π
Solution:

Question 6.
Prove that \(\cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)\) = 7
Solution:

Question 7.
Show that \(\sin ^{-1}\left(\frac{1}{\sqrt{17}}\right)+\cos ^{-1}\left(\frac{9}{\sqrt{85}}\right)=\tan ^{-1}\left(\frac{1}{2}\right)\)
Solution:
Firstly we convert sin^-1 to tan^-1, for this we construct a right triangle with p = 1; h = \(\sqrt{17}\)

Now we convert cos^-1 to tan^-1, for this we construct a right triangle with b = 9; h = \(\sqrt{85}\)

Question 8.
Prove that \(2 \tan ^{-1}\left(\frac{1}{3}\right)+\cot ^{-1}(4)=\tan ^{-1} \frac{16}{13}\)
Solution:

Question 9.
Solve for x : tan^-1(x – 1) + tan^-1x + tan^-1(x + 1) = tan^-13x
Solution:
Given tan^-1(x – 1) + tan^-1x + tan^-1(x + 1) = tan^-13x
⇒ tan^-1(x – 1) + tan^-1(x + 1) = tan^-13x – tan^-1x
L.H.S. = tan^-1(x – 1) + tan^-1(x + 1)

All the values of x satisfies given condition
Thus, x = 0, ± \(\frac { 1 }{ 2 }\)

Question 10.
If sin^-1 x + sin^-1 y + sin^-1 z = π, prove that x² – y² – z² + 2yz\(\sqrt{1-x^2}\) = 0. Solution:
Given sin^-1x + sin^-1y + sin^-1z = π
⇒ \(\sin ^{-1}\left[x \sqrt{1-y^2}+y \sqrt{1-x^2}\right]=\pi-\sin ^{-1} z\)
⇒ \(x \sqrt{1-y^2}+y \sqrt{1-x^2}=\sin \left(\pi-\sin ^{-1} z\right)=\sin \left(\sin ^{-1} z\right)\)
⇒ \(x \sqrt{1-y^2}+y \sqrt{1-x^2}=z\) … (1)
⇒ \(x \sqrt{1-y^2}-z=-y \sqrt{1-x^2}\)
on squaring both sides, we have
x²(1 – y²) + z² – 2xz\(\sqrt{1-y^2}\) = y²(1 – x²)
⇒ x² – x²y² + z² – 2xz\(\sqrt{1-y^2}\) = y² – x²y²
⇒ x² + z² – y² – 2xz\(\sqrt{1-y^2}\) = 0
Also from (1); x\(\sqrt{1-y^2}\) = z – y\(\sqrt{1-x^2}\)
on squaring both sides; we have
x² (1 – y²) = z² + y²(1 – x²) – 2yz\(\sqrt{1-x^2}\)
⇒ x² – x²y² = z² + y² – x²y² – 2yz\(\sqrt{1-x^2}\)
⇒ x² – z² – y² + 2yz\(\sqrt{1-x^2}\) = 0

Question 11.
Solve: \(\cos ^{-1}\left[\sin \left(\cos ^{-1} x\right)\right]=\frac{\pi}{3}\)
Solution:
Given, \(\cos ^{-1}\left[\sin \left(\cos ^{-1} x\right)\right]=\frac{\pi}{3}\) … (1)
put cos^-1x = θ ⇒ x = cos θ,
∴ eqn. (1) becomes; cos^-1(sin θ) = \(\frac { π }{ 3 }\)
⇒ \(\cos ^{-1}\left[\cos \left(\frac{\pi}{2}-\theta\right)\right]=\frac{\pi}{3}\)
⇒ \(\frac{\pi}{2}-\theta=\frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{6}\)
⇒ \(\cos ^{-1} x=\frac{\pi}{6} \Rightarrow x=\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}\)

Question 12.
Prove that \(\sin \left[2 \tan ^{-1} \frac{3}{5}-\sin ^{-1} \frac{7}{25}\right]=\frac{304}{425}\)
Solution:

Question 13.
Solve for x : sin (2 tan^-1 x) = 1
Solution:
Given, sin(2 tan^-1x) = 1
⇒ 2 tan^-1x – sin^-1(1) = sin^-1(sin \(\frac { π }{ 2 }\))
⇒ 2 tan^-1x = \(\frac { π }{ 2 }\)
[∵ sin^-1 (sin x) = x ∀ x ∈ \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\) ]
⇒ 2 tan^-1 x = \(\frac { π }{ 4 }\) ⇒ x = tan \(\frac { π }{ 4 }\) = 1
Clearly, x = 1 satisfies the given eqn.
Hence, x = 1 be the required solution.

Question 14.
Prove that
2\(\tan ^{-1} \frac{1}{5}+\cos ^{-1} \frac{7}{5 \sqrt{2}}+2 \tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\)
Solution:

Question 15.
Prove that
sec² (tan^-12) + cosec² (cot^-13) = 15
Solution:
L.H.S. sec² (tan^-12) + cosec² (cot^-13)
= 1 + tan² (tan^-1 2) + 1 + cot² (cot^-13)
[∵ sec²θ = 1 + tan²θ; cosec²θ = 1 + cot²θ]
= 2+ {tan (tan^-12)}² + {cot (cot^-13)}²
= 2 + 2² + 3² = 2 + 4 + 9 = 15 = R.H.S
[∵ tan(tan²x) = cot(cot^-1x) = x ∀ x ∈ R]

Question 16.
Prove that
\(\cos ^{-1} \frac{63}{65}+2 \tan ^{-1} \frac{1}{5}=\sin ^{-1} \frac{3}{5}\)
Solution:
L.H.S. = \(\cos ^{-1} \frac{63}{65}+2 \tan ^{-1} \frac{1}{5}\)
first of all, we convert cos^-1 to sin^-1, for this we construct right triangle with b = 63 & h = 65

Question 17.
Prove that
\(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \sin ^{-1} \frac{4}{5}\)
Solution:

Question 18.
Solve for x :
sin^-1 x + sin^-1 (1 – x) = cos^-1 x, x ≠ 0
Solution:
Given

Question 19.
Evaluate : tan\(\left[2 \tan ^{-1} \frac{1}{2}-\cot ^{-1} 3\right]\)
Solution:

Question 20.
If cos^-1 x + cos^-1 y + cos^-1 z = π, prove that x² + y² + z² + 2xyz = 1.
Solution:
Given eqn. be,
cos^-1 x + cos^-1 y + cos^-1 z = π
cos^-1 x + cos^-1 y = π – cos^-1 z
⇒ cos^-1\(\left\{x y-\sqrt{1-x^2} \sqrt{1-y^2}\right\}\) = π-cos^-1z
⇒ xy – \(\sqrt{1-x^2} \sqrt{1-y^2}\)
= cos(π – cos^-1z)
⇒ xy – \(\sqrt{1-x^2} \sqrt{1-y^2}\) = – cos(cos^-1z)
⇒ xy – \(\sqrt{1-x^2} \sqrt{1-y^2}\) = – z
[cos(cos^-1x) = x ∀ |x| ≤ 1]
⇒ xy + z = \(\sqrt{1-x^2} \sqrt{1-y^2}\)
on squraing both sides, we have
⇒ (xy + z)² = (1 – x²)(1 – y²)
⇒ x²y² + z² + 2xyz = 1 – x² – y² + x²y²
⇒ x² + y² + z² + 2xyz = 1,
which is the required result.

Question 21.
Solve : \(\cos ^{-1}\left(\sin \cos ^{-1} x\right)=\frac{\pi}{6}\)
Solution:

Question 22.
Solve the equation for x :
\(\sin ^{-1} \frac{5}{x}+\sin ^{-1} \frac{12}{x}=\frac{\pi}{2}, x \neq 0\)
Solution:

Question 23.
Solve for x, if tan(cos^-1 x) = \(\frac{2}{\sqrt{5}}\)
Solution:
Given tan (cos^-1 x) = \(\frac{2}{\sqrt{5}}\)
we convert cos^-1 to tan^-1 for this we construct a right triangle with b = x ; h = 1

Question 24.
If sin^-1 x + tan^-1x = \(\frac { π }{ 2 }\), prove that 2x² + 1 = \(\sqrt{5}\).
Solution:
Given, sin^-1 x + tan^-1x = \(\frac { π }{ 2 }\)

⇒ x = \(\frac{1}{\sqrt{1+x^2}}\)
on squaring both sides, we have
x² = \(\frac{1}{1+x^2}\)
⇒ x²(1 + x²) – 1 = 0
⇒ x⁴ + x² – 1 = 0
⇒ x² = \(\frac{-1 \pm \sqrt{5}}{2}\)
but x² ≥ 0
∴ x² = \(\frac{-1+\sqrt{5}}{2}\)
⇒ 2x² + 1 = \(\sqrt{5}\) which is the required result.

Question 25.
Prove that
\(\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)=\tan ^{-1} \sqrt{x}\)
Solution:
Put x = tan² θ ⇒ \(\sqrt{x}\) = tan θ
⇒ θ = tan^-1\(\sqrt{x}\)
L.H.S = \(\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\)
= \(\frac{1}{2} \cos ^{-1}\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)\)
= \(\frac{1}{2} \cos ^{-1}(\cos 2 \theta)=\frac{1}{2}\) x 2θ
= θ = tan^-1 \(\sqrt{x}\)

Question 26.
\(\sin ^{-1}\left(\sin \frac{\pi}{6}\right)\) = …………..
Solution:
Let sin^-1\(\left(\sin \frac{\pi}{6}\right)=x ;-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\)
⇒ \(\sin \frac{\pi}{6}=\sin x ;-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\)
⇒ x = \(\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
∴ sin^-1\(\left(\sin \frac{\pi}{6}\right)=\frac{\pi}{6}\)

Question 27.
\(\sin ^{-1}\left(\sin \frac{5 \pi}{6}\right)\) = ………….
Solution:

Question 28.
The principal values sec^-1(- 2) is ………….
Solution:

Question 29.
If sin^-1 α = tan^-1\(\frac { 3 }{ 4 }\), then α equals ………….
Solution:
Given sin^-1 α = tan^-1\(\frac { 3 }{ 4 }\)
We convert tan^-1 to sin^-1, for this, we construct a triangle with p = 3 ; b = 4

Question 30.
cos^-1 (- 1) – sin^-1 (1) = ………….
Solution:
cos^-1 (- 1) = cos^-1 (cos π) = π
[∵ cos^-1 (cos θ) = θ if θ ∈ [0, π]]
and \(\sin ^{-1} 1=\sin ^{-1}\left(\sin \frac{\pi}{2}\right)=\frac{\pi}{2}\)
[∵ sin^-1 (sin θ) = θ if θ ∈ \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) ]
∴ \(\cos ^{-1}(-1)-\sin ^{-1}(1)=\pi-\frac{\pi}{2}=\frac{\pi}{2}\)

Question 31.
\(\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}\) = ………….
Solution:

Question 32.
The vale of cos (sin^-1 x) is ………….
Solution:

Question 33.
\(\cos \left[\cos ^{-1}\left(-\frac{1}{9}\right)+\sin ^{-1}\left(-\frac{1}{9}\right)\right]\) is equal to ………….
Solution:
We know that,

Question 34.
3 tan^-1 a is equal to ………….
Solution:

Question 35.
The result tan^-1x – tan^-1y = tan^-1\(\left(\frac{x-y}{1+x y}\right)\) is true when value of xy ………..
Solution:

Question 36.
\(\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\) is equal to
(a) \(\frac { π }{ 4 }\)
(b) \(\frac { π }{ 6 }\)
(c) \(\frac { π }{ 3 }\)
(d) \(\frac { 2π }{ 3 }\)
Solution:

Question 37.
\(\cos ^{-1}\left(-\frac{1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)\) equal
(a) \(\frac { 19π }{ 12 }\)
(b) \(\frac { 13π }{ 12 }\)
(c) \(\frac { 47π }{ 12 }\)
(d) \(\frac { 43π }{ 12 }\)
Solution:

Question 38.
The value of \(\cos ^{-1}\left(\sin \frac{7 \pi}{6}\right)\) is
(a) – \(\frac { π }{ 3 }\)
(b) \(\frac { π }{ 6 }\)
(c) \(\frac { π }{ 3 }\)
(d) \(\frac { 2π }{ 3 }\)
Solution:
\(\cos ^{-1}\left(\sin \frac{7 \pi}{6}\right)=\cos ^{-1}\left(\sin \left(\pi+\frac{\pi}{6}\right)\right)=\cos ^{-1}\left(-\sin \frac{\pi}{6}\right)=\cos ^{-1}\left(-\frac{1}{2}\right)\)
= \(\cos ^{-1}\left(-\cos \left(\frac{\pi}{3}\right)\right)=\cos ^{-1}\left\{\cos \left(\pi-\frac{\pi}{3}\right)\right\}=\cos ^{-1}\left\{\cos \frac{2 \pi}{3}\right\}=\frac{2 \pi}{3}\)
[∵ cos^-1 (cos θ = θ; θ ∈ [θ, π]

Question 39.
The value of \(\sin ^{-1}\left(\cos \frac{53 \pi}{5}\right)\) is
(a) \(\frac { 3π }{ 5 }\)
(b) –\(\frac { 3π }{ 5 }\)
(c) \(\frac { π }{ 10 }\)
(d) –\(\frac { π }{ 10 }\)
Solution:

Question 40.
The value of sin^-1 {cos (4905°)} is equal to
(a) –\(\frac { π }{ 3 }\)
(b) \(\frac { π }{ 6 }\)
(c) –\(\frac { π }{ 4 }\)
(d) \(\frac { π }{ 4 }\)
(e) \(\frac { π }{ 2 }\)
Solution:
sin^-1 {cos (4905°)} = sin^-1 {cos (3960° + 135°)} = sin^-1 {cos (360° x 11 + 135°)}
= sin^-1 {cos 135°} = sin^-1 {cos (90° + 45°)} = sin^-1 {- sin 45°}
= sin^-1 {sin (- 45°)} =
[∵ sin^-1 (sin θ) = θ ; θ ∈ \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) ]

Question 41.
The value of sin [2 sin^-1(cos A)] is
(a) sin A
(b) cos A
(c) sin 2A
(d) cos 2A
Solution:
sin [2 sin^-1(cos A)]
= \(\sin \left[2 \sin ^{-1}\left(\sin \left(\frac{\pi}{2}-\mathrm{A}\right)\right)\right]\)
= \(\sin \left[2\left(\frac{\pi}{2}-\mathrm{A}\right)\right]\)
= sin(π – 2A) = sin 2A

Question 42.
\(\cos \left[2 \cos ^{-1} \frac{1}{5}+\sin ^{-1} \frac{1}{5}\right]\) is equal to
(a) –\(\frac{\sqrt{3}}{2}\)
(b) \(\frac { 4 }{ 5 }\)
(c) –\(\frac{2 \sqrt{6}}{5}\)
(d) \(\frac{2}{3 \sqrt{3}}\)
Solution:

Question 43.
If sin^-1 \(\frac { x }{ 5 }\) + cosec^-1\(\frac { 5 }{ 4 }\) = \(\frac { π }{ 2 }\), then x is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Solution:

Question 44.
tan^-1 1 + tan^-1 2 + tan^-1 3 is
(a) 0
(b) tan^-1 6
(c) \(\frac { π }{ 4 }\)
(d) π
Solution:

Question 45.
The value of cos (2 cos^-1 0.80) is
(a) 0.28
(b) 0.48
(c) 0.84
(d) 0.96
Solution:
\(\cos \left(2 \cos ^{-1} 0.80\right)=\cos \left(2 \cos ^{-1} \frac{4}{5}\right)=\cos \left\{\cos ^{-1}\left(2\left(\frac{4}{5}\right)^2-1\right)\right\}\) [∵ 2 cos^-1 x = cos^-1(2x² – 1)]
= \(\cos \left\{\cos ^{-1}\left(\frac{7}{25}\right)\right\}=\frac{7}{25}\)

Question 46.
The value of cos^-1(cos\(\frac { 3π }{ 2 }\)) is
(a) \(\frac { π }{ 2 }\)
(b) \(\frac { 3π }{ 2 }\)
(c) \(\frac { 5π }{ 2 }\)
(d) \(\frac { 7π }{ 2 }\)
Solution:
\(\cos ^{-1}\left(\cos \frac{3 \pi}{2}\right)=\cos ^{-1}\left\{\cos \left(\pi+\frac{\pi}{2}\right)\right\}=\cos ^{-1}\left\{-\cos \frac{\pi}{2}\right\}=\cos ^{-1} 0=\frac{\pi}{2}\)

Question 47.
If \(\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{3}{4}\right)-\tan ^{-1}\left(\frac{x}{3}\right)\) = 0 is
(a) \(\frac { 7 }{ 3 }\)
(b) 3
(c) \(\frac { 11 }{ 3 }\)
(d) \(\frac { 13 }{ 3 }\)
Solution:

Question 48.
The solution of is \(\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}\) is
(a) \(-\frac{1}{\sqrt{3}}\)
(b) \(\frac{1}{\sqrt{3}}\)
(c) –\(\sqrt{3}\)
(d) \(\sqrt{3}\)
Solution:

Question 49.
If \(\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}\), then the value of cos^-1 x + cos^-1 y is
(a) \(\frac { 2π }{ 3 }\)
(b) \(\frac { π }{ 3 }\)
(c) \(\frac { π }{ 2 }\)
(d) π
Solution:

Question 50.
Find the principal value of \(\sin ^{-1}\left[\cos \left(\sin ^{-1} \frac{1}{2}\right)\right]\).
Solution:

Question 51.
Find the value of tan\(\left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)\).
Solution:
\(\tan \left(\frac{\sin ^{-1} x+\cos ^{-1} x}{2}\right)=\tan \left(\frac{\pi}{4}\right)=1\)
[ ∵ \(\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \forall x \in\) [-1, 1] ]

Question 52.
Find the principal value of cos^-1(-\(\frac { 1 }{ 2 }\)).
Solution:

Question 53.
If sin^-1 \(\frac { x }{ 5 }\) + cosec^-1\(\frac { 5 }{ 4 }\) = \(\frac { π }{ 2 }\), then find x.
Solution:
Given sin^-1 \(\frac { x }{ 5 }\) + cosec^-1\(\frac { 5 }{ 4 }\) = \(\frac { π }{ 2 }\)

On squaring both sides ;
⇒ 16 (25 – x²) = (25 – 3x)²
⇒ 400 – 16x² = 625 + 9x² – 150x
⇒ 25x² – 150x + 225 = 0
⇒ x² – 6x + 9 = 0
⇒ (x – 3)² = 0
⇒ x = 3

Question 54.
Find or if 4 sin^-1 x + cos^-1 x = π.
Solution:
Given 4 sin^-1 x + cos^-1 x = π

Question 55.
Find x if sin^-1 x – cos^-1 x = \(\frac { π }{ 6 }\).
Solution:
Given sin^-1 x – cos^-1 x = \(\frac { π }{ 6 }\)
⇒ \(\sin ^{-1} x-\left(\frac{\pi}{2}-\sin ^{-1} x\right)=\frac{\pi}{6}\)
⇒ \(2 \sin ^{-1} x=\frac{\pi}{6}+\frac{\pi}{2}\)
⇒ \(2 \sin ^{-1} x=\frac{2 \pi}{3} \quad \Rightarrow \sin ^{-1} x=\frac{\pi}{3}\)
⇒ x = \(\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\)

Question 56.
Find the principal values of
(i) sin^-1\(\left(\frac{-1}{\sqrt{2}}\right)\)
(ii) cos^-1\(\frac{\sqrt{3}}{2}\)
(iii) tan^-1\((-\sqrt{3})\)
(iv) cot^-1(-1)
(v) sec^-1\(\frac{2}{\sqrt{3}}\)
(vi) sec^-1(- 2)
(vii) sin^-1\(\left(\frac{-1}{2}\right)\)
(viii) cos^-1\(\left(\frac{-\sqrt{3}}{2}\right)\)
(ix) tan^-1\(\left(\frac{1}{\sqrt{3}}\right)\)
(x) tan^-1\((-\sqrt{3})\)
(xi) sec^-1\((-\sqrt{2})\)
(xii) cot^-1\((-\sqrt{3})\)
Solution:

Question 57.
Evaluate:
\(\tan ^{-1}(\sqrt{3})-\sec ^{-1}(-2)\)+cosec^-1\(\frac{2}{\sqrt{3}}\)
Solution:

Question 58.
Prove that
\(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\)
Solution:

Question 59.
Prove that
\(\sin ^{-1}\left(\frac{8}{17}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\cos ^{-1}\left(\frac{36}{85}\right)\)
Solution:

we convert sin 1 to cos 1 for this, we construct a right triangle with p = 77 & h = 85
∴ b = \(\sqrt{h^2-p^2}=\sqrt{85^2-77^2}\)
= \(\sqrt{(85-77)(85+77)}=\sqrt{8 \times 162}\)
= \(\sqrt{1296}\) = 36
Thus \(\sin ^{-1}\left(\frac{77}{85}\right) \cos ^{-1}\left(\frac{36}{85}\right)\)

Question 60.
Prove that
\(\tan ^{-1} \frac{1}{2}=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1}\left(\frac{4}{5}\right)\)
Solution:
We want to prove that,

Question 61.
Prove that
\(\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\)
Solution: