CBSE Class 12 Mathematics Inverse Trigonometric Functions Worksheet Set 01

CBSE Class 12 Mathematics Inverse Trigonometric Functions (1). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

MULTIPLE CHOICE QUESTIONS

Question. tan^-1 √3 – sec^-1 (-2) is equal to
(a) π
(b) –π/3
(c) π/3
(d) 2π/3
Answer : B

Question. Principal value of tan^-1 (-1) is
(a) π/4
(b) −π/2
(c) 5π/4
(d) −π/4
Answer : D

Question. The principle value of sin^-1(sin2π/3) is
(a) 2π/3
(b) π/3
(c) −π/6
(d) π/6
Answer : B

Question. Simplified form of cos^-1 (4x3 – 3x)
(a) 3 sin^-1x
(b) 3 cos^-1x
(c) π – 3 sin^-1x
(d) None of these
Answer : B

Question. cos^-1(cos 7π/6) is equal to
(a) 7π/6
(b) 5π/6
(c) π/3
(d) π/6
Answer : B

Question. The value of cos^-1(1/2) + 2sin^-1(1/2) is equal to
(a) π/4
(b) π/6
(c) 2π/3
(d) 5π/6
Answer : B

Question. sin^-1 x = y Then
(a) 0 ≤ y ≤ π
(b) –π/2 ≤ y ≤ π/2
(c) 0 < y < π
(d) –π/2 < y < –π/2
Answer : B

Question. Principal value of sin^-1(1/√2)
(a) π/4
(b) 3π/4
(c) 5π/4
(d) None of these
Answer : A

Question. The principal value of cosec^-1 (-2) is
(a) –2π/3
(b) π/6
(c) 2π/3
(d) –π/6
Answer : D

Question. The value of expression 2 sec^-1 (2) + sin^-1 (1/2) is
(a) π/6
(b) 5π/6
(c) 7π/6
(d) 1
Answer : B

Question. The principle value of sin^-1 (√3/2) is
(a) 2π/3
(b) π/6
(c) π/4
(d) π/3
Answer : D

Question. sin[π/3 – sin^-1(-1/2)] is equal to
(a) 1//2
(b) 1/3
(c) 1/4
(d) 1
Answer : D

CASE STUDY QUESTIONS

Case Study 1

A group of students of class XII visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Raj path (formerly called the Kingsway), is about 138 feet (42 metrs) in height.

Question. What is the angle of elevation if they are standing at a distance of 42m away from the monument?
a) tan⁻¹ 1
b) sin⁻¹ 1
c) cos⁻¹ 1
d) sec⁻¹ 1
Answer : A

Question. They want to see the tower at an angle of sec⁻¹ 1/2. So, they want to know the distance where they should stand and hence find the distance.
a) 42 m
b) 20.12 m
c) 25.24 m
d) 24.64 m
Answer : C

Question. If the altitude of the Sun is at cos−1 1/2, then the height of the vertical tower that will cast a shadow of length 20 m is
a) 20√3 m
b) 20/ √3 m
c) 15/ √3 m
d) 15√3 m
Answer : A

Question. The ratio of the length of a rod and its shadow is 1:2. The angle of elevation of the Sun is
a) sin⁻¹ 1/2
b) cos⁻¹ 1/2
c) tan⁻¹ 1/2
d) cot⁻¹ 1/2
Answer : A

Question. Domain of sin⁻¹ 𝑥 is……..
a) (-1, 1)
b) {-1,1}
c) [ -1,1]
d) none of these
Answer : C

Case Study 2

A Satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka, them being Nanda Devi (height 7,816m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are cot−1 √3 andtan−1 √3 respectively. If the distance between the peaks of the two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains.

Question. The distance of the satellite from the top of Nanda Devi is
a) 1139.4 km
b) 577.52 km
c) 1937 km
d) 1025.36 km
Answer : A

Question. The distance of the satellite from the top of Mullayanagiri is
a) 1139.4 km
b) 577.52 km
c) 1937 km
d) 1025.36 km
Answer : C

Question. The distance of the satellite from the ground is
a) 1139.4 km
b) 577.52 km
c) 1937 km
d) 1025.36 km
Answer : A

Question. What is the angle of elevation if a man is standing at a distance of 7816m from Nanda Devi?
a) sec⁻¹ 2
b) cot⁻¹ 1
c) sin⁻¹ √3
d) cos⁻¹ 1/2
Answer : A

ONE MARK QUESTIONS

Question. Find the principal value of the following:
a) \( \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \)
b) \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) \)
c) \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)
d) \( \csc^{-1}(-2) \)
e) \( \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) \)
Answer:
a) Let \( \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \theta \). Since the principal value branch of \( \sin^{-1} \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \), we have:
\( \sin\theta = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4} \)

b) Let \( \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \theta \). Since the principal value branch of \( \cos^{-1} \) is \( [0, \pi] \), we have:
\( \cos\theta = \frac{\sqrt{3}}{2} \implies \theta = \frac{\pi}{6} \)

c) Let \( \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = \theta \). Since the principal value branch of \( \tan^{-1} \) is \( (-\frac{\pi}{2}, \frac{\pi}{2}) \), we have:
\( \tan\theta = -\frac{1}{\sqrt{3}} \implies \theta = -\frac{\pi}{6} \)

d) Let \( \csc^{-1}(-2) = \theta \). Since the principal value branch of \( \csc^{-1} \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\} \), we have:
\( \csc\theta = -2 \implies \sin\theta = -\frac{1}{2} \implies \theta = -\frac{\pi}{6} \)

e) Let \( \sec^{-1}\left(-\frac{2}{\sqrt{3}}\right) = \theta \). Since the principal value branch of \( \sec^{-1} \) is \( [0, \pi] - \{\frac{\pi}{2}\} \), we have:
\( \sec\theta = -\frac{2}{\sqrt{3}} \implies \cos\theta = -\frac{\sqrt{3}}{2} \implies \theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6} \)

Question. Find the value of the following:
a) \( \sin^{-1}\left(\sin \frac{3\pi}{5}\right) \)
b) \( \cos^{-1}\left(\cos \frac{13\pi}{6}\right) \)
c) \( \tan^{-1}\left(\tan \frac{7\pi}{6}\right) \)
d) \( \csc^{-1}\left(\csc \frac{\pi}{8}\right) \)
e) \( \sec^{-1}\left(\sec \frac{3\pi}{4}\right) \)
Answer:
a) \( \sin^{-1}\left(\sin \frac{3\pi}{5}\right) = \sin^{-1}\left(\sin\left(\pi - \frac{2\pi}{5}\right)\right) = \sin^{-1}\left(\sin \frac{2\pi}{5}\right) = \frac{2\pi}{5} \) (since \( \frac{2\pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}] \)).

b) \( \cos^{-1}\left(\cos \frac{13\pi}{6}\right) = \cos^{-1}\left(\cos\left(2\pi + \frac{\pi}{6}\right)\right) = \cos^{-1}\left(\cos \frac{\pi}{6}\right) = \frac{\pi}{6} \) (since \( \frac{\pi}{6} \in [0, \pi] \)).

c) \( \tan^{-1}\left(\tan \frac{7\pi}{6}\right) = \tan^{-1}\left(\tan\left(\pi + \frac{\pi}{6}\right)\right) = \tan^{-1}\left(\tan \frac{\pi}{6}\right) = \frac{\pi}{6} \) (since \( \frac{\pi}{6} \in (-\frac{\pi}{2}, \frac{\pi}{2}) \)).

d) \( \csc^{-1}\left(\csc \frac{\pi}{8}\right) = \frac{\pi}{8} \) (since \( \frac{\pi}{8} \in [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\} \)).

e) \( \sec^{-1}\left(\sec \frac{3\pi}{4}\right) = \frac{3\pi}{4} \) (since \( \frac{3\pi}{4} \in [0, \pi] - \{\frac{\pi}{2}\} \)).

Question. Evaluate the following:
a) \( \sin\left\{ \frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right) \right\} \)
b) \( \sin\left( \frac{1}{2}\cos^{-1}\frac{1}{9} \right) \)
c) \( \tan\left[ \frac{1}{2}\cos^{-1}\left(\frac{\sqrt{5}}{3}\right) \right] \)
Answer:
a) \( \sin\left\{ \frac{\pi}{3} - \sin^{-1}\left(-\frac{1}{2}\right) \right\} = \sin\left\{ \frac{\pi}{3} - \left(-\frac{\pi}{6}\right) \right\} = \sin\left( \frac{\pi}{3} + \frac{\pi}{6} \right) = \sin\left(\frac{\pi}{2}\right) = 1 \).

b) Let \( \cos^{-1}\frac{1}{9} = \theta \implies \cos\theta = \frac{1}{9} \).
We know that:
\( \sin\frac{\theta}{2} = \sqrt{\frac{1 - \cos\theta}{2}} = \sqrt{\frac{1 - \frac{1}{9}}{2}} = \sqrt{\frac{8}{18}} = \sqrt{\frac{4}{9}} = \frac{2}{3} \).

c) Let \( \cos^{-1}\left(\frac{\sqrt{5}}{3}\right) = \theta \implies \cos\theta = \frac{\sqrt{5}}{3} \).
We need to find:
\( \tan\frac{\theta}{2} = \sqrt{\frac{1-\cos\theta}{1+\cos\theta}} = \sqrt{\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}} = \sqrt{\frac{(3-\sqrt{5})^2}{9-5}} = \frac{3-\sqrt{5}}{2} \).

Question. Evaluate: \( \cos\left( \sin^{-1}\frac{3}{5} + \cos^{-1}\frac{12}{13} \right) \)
Answer: Let \( A = \sin^{-1}\frac{3}{5} \implies \sin A = \frac{3}{5} \) and \( \cos A = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5} \).
Let \( B = \cos^{-1}\frac{12}{13} \implies \cos B = \frac{12}{13} \) and \( \sin B = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13} \).
We need to find:
\( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
\( = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \)
\( = \frac{48}{65} - \frac{15}{65} = \frac{33}{65} \).

Question. Show that \( \tan^{-1}(\sqrt{x}) = \frac{1}{2}\cos^{-1}\left( \frac{1-x}{1+x} \right) \).
Answer: Let \( \tan^{-1}\sqrt{x} = \theta \implies \tan\theta = \sqrt{x} \implies \tan^2\theta = x \).
Now, let's consider the Right Hand Side (R.H.S):
\( \text{R.H.S} = \frac{1}{2}\cos^{-1}\left( \frac{1-x}{1+x} \right) \)
Substituting \( x = \tan^2\theta \):
\( = \frac{1}{2}\cos^{-1}\left( \frac{1-\tan^2\theta}{1+\tan^2\theta} \right) \)
Since \( \cos(2\theta) = \frac{1-\tan^2\theta}{1+\tan^2\theta} \), we have:
\( = \frac{1}{2}\cos^{-1}(\cos(2\theta)) \)
\( = \frac{1}{2}(2\theta) \)
\( = \theta \)
\( = \tan^{-1}\sqrt{x} = \text{L.H.S} \).
Hence proved.

FOUR MARKS QUESTIONS

Question. Prove that \( \tan^{-1}\left( \frac{\sqrt{1+x^2} + \sqrt{1-x^2}}{\sqrt{1+x^2} - \sqrt{1-x^2}} \right) = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}x^2 \).
Answer: Let \( x^2 = \cos(2\theta) \implies 2\theta = \cos^{-1}x^2 \implies \theta = \frac{1}{2}\cos^{-1}x^2 \).
We know that:
\( 1 + x^2 = 1 + \cos(2\theta) = 2\cos^2\theta \implies \sqrt{1+x^2} = \sqrt{2}\cos\theta \)
\( 1 - x^2 = 1 - \cos(2\theta) = 2\sin^2\theta \implies \sqrt{1-x^2} = \sqrt{2}\sin\theta \)
Substituting these into the Left Hand Side (L.H.S):
\( \text{L.H.S} = \tan^{-1}\left( \frac{\sqrt{2}\cos\theta + \sqrt{2}\sin\theta}{\sqrt{2}\cos\theta - \sqrt{2}\sin\theta} \right) \)
\( = \tan^{-1}\left( \frac{\cos\theta + \sin\theta}{\cos\theta - \sin\theta} \right) \)
Dividing the numerator and denominator by \( \cos\theta \):
\( = \tan^{-1}\left( \frac{1 + \tan\theta}{1 - \tan\theta} \right) \)
\( = \tan^{-1}\left( \tan\left( \frac{\pi}{4} + \theta \right) \right) \)
\( = \frac{\pi}{4} + \theta \)
Substituting back \( \theta = \frac{1}{2}\cos^{-1}x^2 \):
\( = \frac{\pi}{4} + \frac{1}{2}\cos^{-1}x^2 = \text{R.H.S} \).
Hence proved.

Question. Prove that \( \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} = \frac{1}{2}\cos^{-1}\frac{3}{5} \).
Answer: Let's evaluate the Left Hand Side (L.H.S):
\( \text{L.H.S} = \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} \)
Using the formula \( \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left( \frac{a+b}{1-ab} \right) \) (since \( ab = \frac{2}{36} < 1 \)):
\( = \tan^{-1}\left( \frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{1}{4}\cdot\frac{2}{9}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{17}{36}}{\frac{34}{36}} \right) \)
\( = \tan^{-1}\left( \frac{17}{34} \right) = \tan^{-1}\frac{1}{2} \)
Let \( \tan^{-1}\frac{1}{2} = \theta \implies \tan\theta = \frac{1}{2} \).
We know that:
\( \cos(2\theta) = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \frac{1 - \left(\frac{1}{2}\right)^2}{1 + \left(\frac{1}{2}\right)^2} = \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5} \)
\( \Rightarrow 2\theta = \cos^{-1}\frac{3}{5} \implies \theta = \frac{1}{2}\cos^{-1}\frac{3}{5} \).
Therefore,
\( \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} = \frac{1}{2}\cos^{-1}\frac{3}{5} \).
Hence proved.

Question. Prove that \( \cot^{-1}\left(\frac{ab+1}{a-b}\right) + \cot^{-1}\left(\frac{bc+1}{b-c}\right) + \cot^{-1}\left(\frac{ca+1}{c-a}\right) = 0 \). (NCERT EXEMPLAR)
Answer: We know that \( \cot^{-1}\left(\frac{xy+1}{y-x}\right) = \tan^{-1}y - \tan^{-1}x \) (for \( x, y > 0 \)).
Let's rewrite each term using this property:
\( \cot^{-1}\left(\frac{ab+1}{a-b}\right) = \tan^{-1}a - \tan^{-1}b \)
\( \cot^{-1}\left(\frac{bc+1}{b-c}\right) = \tan^{-1}b - \tan^{-1}c \)
\( \cot^{-1}\left(\frac{ca+1}{c-a}\right) = \tan^{-1}c - \tan^{-1}a \)
Adding these three terms together:
\( \text{L.H.S} = (\tan^{-1}a - \tan^{-1}b) + (\tan^{-1}b - \tan^{-1}c) + (\tan^{-1}c - \tan^{-1}a) \)
\( = 0 = \text{R.H.S} \).
Hence proved.

Question. Prove that \( \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{3}{5} - \tan^{-1}\frac{8}{19} = \frac{\pi}{4} \).
Answer: Let's first combine the first two terms of the Left Hand Side (L.H.S):
\( \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{3}{5} = \tan^{-1}\left( \frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{3}{4}\cdot\frac{3}{5}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{27}{20}}{\frac{11}{20}} \right) = \tan^{-1}\frac{27}{11} \)
Now, let's subtract the third term:
\( \text{L.H.S} = \tan^{-1}\frac{27}{11} - \tan^{-1}\frac{8}{19} \)
Using the formula \( \tan^{-1}x - \tan^{-1}y = \tan^{-1}\left( \frac{x-y}{1+xy} \right) \):
\( = \tan^{-1}\left( \frac{\frac{27}{11} - \frac{8}{19}}{1 + \frac{27}{11}\cdot\frac{8}{19}} \right) \)
\( = \tan^{-1}\left( \frac{27 \times 19 - 8 \times 11}{11 \times 19 + 27 \times 8} \right) \)
Since \( 27 \times 19 - 8 \times 11 = 513 - 88 = 425 \) and \( 11 \times 19 + 27 \times 8 = 209 + 216 = 425 \), we have:
\( = \tan^{-1}\left( \frac{425}{425} \right) = \tan^{-1}(1) = \frac{\pi}{4} = \text{R.H.S} \).
Hence proved.

Question. Prove that \( \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65} \).
Answer: Let \( A = \cos^{-1}\frac{4}{5} \implies \cos A = \frac{4}{5} \) and \( \sin A = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \frac{3}{5} \).
Let \( B = \cos^{-1}\frac{12}{13} \implies \cos B = \frac{12}{13} \) and \( \sin B = \sqrt{1 - \left(\frac{12}{13}\right)^2} = \frac{5}{13} \).
Now, let's consider:
\( \cos(A+B) = \cos A \cos B - \sin A \sin B \)
\( = \left(\frac{4}{5}\right)\left(\frac{12}{13}\right) - \left(\frac{3}{5}\right)\left(\frac{5}{13}\right) \)
\( = \frac{48}{65} - \frac{15}{65} = \frac{33}{65} \)
\( \Rightarrow A + B = \cos^{-1}\frac{33}{65} \)
Substituting back \( A \) and \( B \):
\( \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65} \).
Hence proved.

Question. Solve for x: \( \sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2} \).
Answer: Given equation is:
\( \sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2} \)
\( \Rightarrow \sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}x \)
Taking sine on both sides:
\( 1 - x = \sin\left( \frac{\pi}{2} + 2\sin^{-1}x \right) \)
Since \( \sin\left(\frac{\pi}{2} + \theta\right) = \cos\theta \), we have:
\( 1 - x = \cos(2\sin^{-1}x) \)
We know that \( \cos(2\theta) = 1 - 2\sin^2\theta \). Let \( \theta = \sin^{-1}x \):
\( 1 - x = 1 - 2\left[\sin(\sin^{-1}x)\right]^2 \)
\( \Rightarrow 1 - x = 1 - 2x^2 \)
\( \Rightarrow 2x^2 - x = 0 \)
\( \Rightarrow x(2x - 1) = 0 \)
\( \Rightarrow x = 0 \) or \( x = \frac{1}{2} \).

Let us verify both values in the original equation:
- For \( x = 0 \):
\( \text{L.H.S} = \sin^{-1}(1) - 2\sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2} \) (Satisfied)
- For \( x = \frac{1}{2} \):
\( \text{L.H.S} = \sin^{-1}\left(\frac{1}{2}\right) - 2\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} - 2\left(\frac{\pi}{6}\right) = -\frac{\pi}{6} \neq \frac{\pi}{2} \) (Not satisfied)

Therefore, the only solution is \( x = 0 \).

Question. If \( \cos^{-1}\frac{x}{a} + \cos^{-1}\frac{y}{b} = \theta \), then prove that \( \frac{x^2}{a^2} - \frac{2xy}{ab}\cos\theta + \frac{y^2}{b^2} = \sin^2\theta \). (NCERT EXEMPLAR)
Answer: We know that \( \cos^{-1}A + \cos^{-1}B = \cos^{-1}\left( AB - \sqrt{1-A^2}\sqrt{1-B^2} \right) \).
Using this identity:
\( \cos^{-1}\left( \frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} \right) = \theta \)
\( \Rightarrow \frac{xy}{ab} - \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} = \cos\theta \)
\( \Rightarrow \frac{xy}{ab} - \cos\theta = \sqrt{1 - \frac{x^2}{a^2}}\sqrt{1 - \frac{y^2}{b^2}} \)
Squaring both sides:
\( \left( \frac{xy}{ab} - \cos\theta \right)^2 = \left( 1 - \frac{x^2}{a^2} \right)\left( 1 - \frac{y^2}{b^2} \right) \)
\( \Rightarrow \frac{x^2 y^2}{a^2 b^2} - \frac{2xy}{ab}\cos\theta + \cos^2\theta = 1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} + \frac{x^2 y^2}{a^2 b^2} \)
Canceling \( \frac{x^2 y^2}{a^2 b^2} \) from both sides and rearranging:
\( \frac{x^2}{a^2} - \frac{2xy}{ab}\cos\theta + \frac{y^2}{b^2} = 1 - \cos^2\theta \)
\( \Rightarrow \frac{x^2}{a^2} - \frac{2xy}{ab}\cos\theta + \frac{y^2}{b^2} = \sin^2\theta \).
Hence proved.

Question. Prove that \( \tan\left( \frac{\pi}{4} + \frac{1}{2}\cos^{-1}\frac{a}{b} \right) + \tan\left( \frac{\pi}{4} - \frac{1}{2}\cos^{-1}\frac{a}{b} \right) = \frac{2b}{a} \). (CBSE 2010, 2013)
Answer: Let \( \frac{1}{2}\cos^{-1}\frac{a}{b} = \theta \implies \cos^{-1}\frac{a}{b} = 2\theta \implies \cos(2\theta) = \frac{a}{b} \).
The Left Hand Side (L.H.S) becomes:
\( \text{L.H.S} = \tan\left( \frac{\pi}{4} + \theta \right) + \tan\left( \frac{\pi}{4} - \theta \right) \)
\( = \frac{1 + \tan\theta}{1 - \tan\theta} + \frac{1 - \tan\theta}{1 + \tan\theta} \)
\( = \frac{(1 + \tan\theta)^2 + (1 - \tan\theta)^2}{1 - \tan^2\theta} \)
\( = \frac{2(1 + \tan^2\theta)}{1 - \tan^2\theta} \)
\( = 2 \left( \frac{1 + \tan^2\theta}{1 - \tan^2\theta} \right) \)
We know that \( \cos(2\theta) = \frac{1 - \tan^2\theta}{1 + \tan^2\theta} \), so:
\( = \frac{2}{\cos(2\theta)} \)
Substituting \( \cos(2\theta) = \frac{a}{b} \):
\( = \frac{2}{\frac{a}{b}} = \frac{2b}{a} = \text{R.H.S} \).
Hence proved.

Question. Solve for x: \( \cos^{-1}\left( \frac{x^2 - 1}{x^2 + 1} \right) + \tan^{-1}\left( \frac{2x}{x^2 - 1} \right) = \frac{2\pi}{3} \).
Answer: Let \( x = \tan\theta \). Then:
\( \cos^{-1}\left( \frac{x^2 - 1}{x^2 + 1} \right) = \cos^{-1}\left( -\frac{1 - \tan^2\theta}{1 + \tan^2\theta} \right) = \cos^{-1}(-\cos 2\theta) = \pi - \cos^{-1}(\cos 2\theta) = \pi - 2\theta \)
And:
\( \tan^{-1}\left( \frac{2x}{x^2 - 1} \right) = \tan^{-1}\left( -\frac{2\tan\theta}{1 - \tan^2\theta} \right) = \tan^{-1}(-\tan 2\theta) = -2\theta \)
Substituting these into the given equation:
\( (\pi - 2\theta) + (-2\theta) = \frac{2\pi}{3} \)
\( \Rightarrow \pi - 4\theta = \frac{2\pi}{3} \)
\( \Rightarrow 4\theta = \pi - \frac{2\pi}{3} = \frac{\pi}{3} \)
\( \Rightarrow \theta = \frac{\pi}{12} \)
Therefore:
\( x = \tan\left( \frac{\pi}{12} \right) = 2 - \sqrt{3} \).

Question. Solve for x: \( \tan^{-1}\left( \frac{x-1}{x-2} \right) + \tan^{-1}\left( \frac{x+1}{x+2} \right) = \frac{\pi}{4} \).
Answer: Given equation is:
\( \tan^{-1}\left( \frac{x-1}{x-2} \right) + \tan^{-1}\left( \frac{x+1}{x+2} \right) = \frac{\pi}{4} \)
Using the formula \( \tan^{-1}A + \tan^{-1}B = \tan^{-1}\left( \frac{A+B}{1-AB} \right) \):
\( \tan^{-1}\left[ \frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \left(\frac{x-1}{x-2}\right)\left(\frac{x+1}{x+2}\right)} \right] = \frac{\pi}{4} \)
Taking tangent on both sides:
\( \frac{\frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2) - (x-1)(x+1)}{(x-2)(x+2)}} = 1 \)
\( \Rightarrow \frac{(x^2 + x - 2) + (x^2 - x - 2)}{(x^2 - 4) - (x^2 - 1)} = 1 \)
\( \Rightarrow \frac{2x^2 - 4}{-3} = 1 \)
\( \Rightarrow 2x^2 - 4 = -3 \)
\( \Rightarrow 2x^2 = 1 \)
\( \Rightarrow x^2 = \frac{1}{2} \)
\( \Rightarrow x = \pm\frac{1}{\sqrt{2}} \).