CBSE Class 12 Mathematics HOTs Inverse Trigonometric Functions Set 02

Question. If \(\sin\left(\sin^{-1}\frac{1}{5} + \cos^{-1}x\right) = 1\), then find the value of \(x\).
Answer: Given: \(\sin\left(\sin^{-1}\frac{1}{5} + \cos^{-1}x\right) = 1\)
\(\Rightarrow \sin^{-1}\frac{1}{5} + \cos^{-1}x = \sin^{-1}1\)
\(\Rightarrow \sin^{-1}\frac{1}{5} + \cos^{-1}x = \frac{\pi}{2}\)
\(\because \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\)
\(\Rightarrow \sin^{-1}\frac{1}{5} = \sin^{-1}x \Rightarrow x = \frac{1}{5}\)

Question. If \(\tan^{-1}x + \tan^{-1}y = \frac{\pi}{4}\), \(xy < 1\), then write the value of \(x + y + xy\).
Answer: Given: \(\tan^{-1}x + \tan^{-1}y = \frac{\pi}{4}\) (\(xy < 1\))
\(\Rightarrow \tan^{-1}\left(\frac{x+y}{1-xy}\right) = \frac{\pi}{4}\)
\(\Rightarrow \frac{x+y}{1-xy} = \tan\frac{\pi}{4} = 1\)
\(\Rightarrow x + y = 1 - xy \Rightarrow x + y + xy = 1\)

Question. Write the value of \(\tan\left(2\tan^{-1}\frac{1}{5}\right)\).
Answer: Since \(2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)\), for \(|x| < 1\)
So, \(2\tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{2 \times \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2}\right) = \tan^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right) = \tan^{-1}\left(\frac{5}{12}\right)\)
\(\therefore \tan\left(2\tan^{-1}\frac{1}{5}\right) = \tan\left(\tan^{-1}\frac{5}{12}\right) = \frac{5}{12}\)

Question. Write the principal value of \(\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})\).
Answer: \(\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})\)
\(= \tan^{-1}(\sqrt{3}) - [\pi - \cot^{-1}(\sqrt{3})]\)
\(= \tan^{-1}(\sqrt{3}) + \cot^{-1}(\sqrt{3}) - \pi\)
\(= \frac{\pi}{2} - \pi = -\frac{\pi}{2}\)
\(\therefore\) Required principal value is \(-\frac{\pi}{2}\)

Question. Evaluate \(\sin^{-1}\left(\sin \frac{3\pi}{5}\right)\).
Answer: We know that, \(\sin^{-1}(\sin x) = x\) for \(x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
But \(\frac{3\pi}{5} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), which is the principal value branch of \(\sin^{-1}x\).
So, \(\sin^{-1}\left(\sin \frac{3\pi}{5}\right) = \sin^{-1}\left[\sin \left(\pi - \frac{2\pi}{5}\right)\right]\)
\(= \sin^{-1}\left[\sin \left(\frac{2\pi}{5}\right)\right] = \frac{2\pi}{5}\) and \(\frac{2\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
\(\therefore \sin^{-1}\left(\sin \frac{3\pi}{5}\right) = \frac{2\pi}{5}\)

Question. Find the principal value of \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2)\).
Answer: \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = \tan^{-1}\sqrt{3} - \left(\pi - \sec^{-1}2\right)\)
\(= \frac{\pi}{3} - \left(\pi - \frac{\pi}{3}\right) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}\)
\(\therefore\) Principal value of \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2) = -\frac{\pi}{3}\)

Question. Find the value of \(\tan^{-1}\left(\tan \frac{3\pi}{4}\right)\).
Answer: \(\tan^{-1}\left(\tan \frac{3\pi}{4}\right) \neq \frac{3\pi}{4}\) as the principal value branch of \(\tan^{-1} \theta\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\),
So, \(\tan^{-1}\left(\tan \frac{3\pi}{4}\right) = \tan^{-1}\left[\tan \left(\pi - \frac{\pi}{4}\right)\right]\)
\(= \tan^{-1}\left[-\tan\left(\frac{\pi}{4}\right)\right]\)
\(= \tan^{-1}\left[\tan\left(-\frac{\pi}{4}\right)\right]\) \([\because -\tan\theta = \tan(-\theta)]\)
\(= -\frac{\pi}{4} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
Hence, \(\tan^{-1}\left(\tan \frac{3\pi}{4}\right) = -\frac{\pi}{4}\)

Question. Write the principal value of \(\cos^{-1}\left(\cos \frac{7\pi}{6}\right)\).
Answer: \(\cos^{-1}\left(\cos \frac{7\pi}{6}\right) \neq \frac{7\pi}{6}\) as principal value branch of \(\cos^{-1} \theta\) is \([0, \pi]\).
So, \(\cos^{-1}\left(\cos \frac{7\pi}{6}\right) = \cos^{-1}\left[\cos \left(\pi + \frac{\pi}{6}\right)\right]\)
\(= \cos^{-1}\left[-\cos\left(\frac{\pi}{6}\right)\right] = \cos^{-1}\left[\cos\left(\pi - \frac{\pi}{6}\right)\right] = \cos^{-1}\left(\cos \frac{5\pi}{6}\right) = \frac{5\pi}{6}\)
where \(\frac{5\pi}{6} \in [0, \pi]\).
Hence, \(\cos^{-1}\left(\cos \frac{7\pi}{6}\right) = \frac{5\pi}{6}\)

Question. Using principal values, evaluate the following: \(\cos^{-1}\left(\cos \frac{2\pi}{3}\right) + \sin^{-1}\left(\sin \frac{2\pi}{3}\right)\).
Answer: We know that the range of principal value branch of \(\cos^{-1}\theta\) is \([0, \pi]\) and \(\sin^{-1}\theta\) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Then, \(\cos^{-1}\left(\cos \frac{2\pi}{3}\right) + \sin^{-1}\left(\sin \frac{2\pi}{3}\right)\)
\(= \frac{2\pi}{3} + \sin^{-1}\left[\sin \left(\pi - \frac{\pi}{3}\right)\right]\)
\(= \frac{2\pi}{3} + \sin^{-1}\left(\sin \frac{\pi}{3}\right)\)
\(= \frac{2\pi}{3} + \frac{\pi}{3} = \pi\)

Question. Find the principal value of \(\sin^{-1}\left(\sin \frac{4\pi}{5}\right)\).
Answer: We know that, \(\sin^{-1}(\sin x) = x\).
Therefore, \(\sin^{-1}\left(\sin \frac{4\pi}{5}\right) = \frac{4\pi}{5}\).
But \(\frac{4\pi}{5} \notin \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), which is the principal value branch of \(\sin^{-1}x\).
So, \(\sin\frac{4\pi}{5} = \sin\left(\pi - \frac{\pi}{5}\right) = \sin\frac{\pi}{5}\) and \(\frac{\pi}{5} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Therefore, \(\sin^{-1}\left(\sin \frac{4\pi}{5}\right) = \frac{\pi}{5}\).

Question. If \(\sin^{-1}\left(\frac{1}{3}\right) + \cos^{-1}(x) = \frac{\pi}{2}\), then find \(x\).
Answer: \(\sin^{-1}\left(\frac{1}{3}\right) + \cos^{-1}(x) = \frac{\pi}{2}\)
\(\Rightarrow \sin^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} - \cos^{-1}x \Rightarrow \sin^{-1}\left(\frac{1}{3}\right) = \sin^{-1}x\)
\(\because \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\)
\(\Rightarrow x = \frac{1}{3}\)

Question. If \(\sin^{-1}(x) + \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2}\), then find \(x\).
Answer: \(\sin^{-1}x + \cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{2}\)
\(\Rightarrow \sin^{-1}x + \frac{\pi}{3} = \frac{\pi}{2} \Rightarrow \sin^{-1}x = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}\)
\(\Rightarrow x = \sin\frac{\pi}{6} \Rightarrow x = \frac{1}{2}\)

Question. Using principal value, find the value of \(\cos^{-1}\left(\cos \frac{13\pi}{6}\right)\).
Answer: \(\cos^{-1}\left(\cos \frac{13\pi}{6}\right) \neq \frac{13\pi}{6}\) as the range of principal value branch of \(\cos^{-1} \theta\) is \([0, \pi]\).
So, \(\cos^{-1}\left(\cos \frac{13\pi}{6}\right) = \cos^{-1}\left[\cos \left(2\pi + \frac{\pi}{6}\right)\right]\)
\(= \cos^{-1}\left(\cos \frac{\pi}{6}\right) = \frac{\pi}{6}\)
\(\therefore \cos^{-1}\left(\cos \frac{13\pi}{6}\right) = \frac{\pi}{6}\)

Question. If \(\tan^{-1}(\sqrt{3}) + \cot^{-1}(x) = \frac{\pi}{2}\), then find \(x\).
Answer: \(\tan^{-1}(\sqrt{3}) + \cot^{-1}(x) = \frac{\pi}{2}\)
\(\Rightarrow \frac{\pi}{3} + \cot^{-1}x = \frac{\pi}{2} \Rightarrow \cot^{-1}x = \frac{\pi}{2} - \frac{\pi}{3} = \frac{\pi}{6}\)
\(\Rightarrow x = \cot\frac{\pi}{6} \Rightarrow x = \sqrt{3}\)

Question. Show that, \(\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x\).
Answer: L.H.S. = \(\sin^{-1}\left(2x\sqrt{1-x^2}\right)\)
Putting \(x = \sin\theta\), we get:
L.H.S. = \(\sin^{-1}\left(2\sin\theta\sqrt{1-\sin^2\theta}\right)\)
\(= \sin^{-1}(2\sin\theta\cos\theta)\)
\(= \sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x = \text{R.H.S.}\)

Question. Write into the simplest form: \(\tan^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)\).
Answer: Let \(y = \tan^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)\)
Substituting \(1 + \sin x = \left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2\) and \(1 - \sin x = \left(\sin\frac{x}{2} - \cos\frac{x}{2}\right)^2\):
\[ y = \tan^{-1}\left[ \frac{\sqrt{\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2} + \sqrt{\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)^2}}{\sqrt{\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2} - \sqrt{\left(\sin\frac{x}{2}-\cos\frac{x}{2}\right)^2}} \right] \]
\[ \Rightarrow y = \tan^{-1}\left[ \frac{\left(\sin\frac{x}{2} + \cos\frac{x}{2}\right) + \left(\sin\frac{x}{2} - \cos\frac{x}{2}\right)}{\left(\sin\frac{x}{2} + \cos\frac{x}{2}\right) - \left(\sin\frac{x}{2} - \cos\frac{x}{2}\right)} \right] \]
\[ \Rightarrow y = \tan^{-1}\left[ \frac{2\sin\frac{x}{2}}{2\cos\frac{x}{2}} \right] \]
\[ \Rightarrow y = \tan^{-1}\left[ \tan \frac{x}{2} \right] \Rightarrow y = \frac{x}{2} \]

Question. Solve for \(x\) : \(2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)\).
Answer: \(2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)\)
\(\Rightarrow 2\tan^{-1}(\cos x) - \tan^{-1}(2\csc x) = 0\)
\(\Rightarrow \tan^{-1}\left(\frac{2\cos x}{1-\cos^2x}\right) - \tan^{-1}(2\csc x) = 0\)
\(\Rightarrow \tan^{-1}\left(\frac{2\cos x}{\sin^2x}\right) - \tan^{-1}\left(\frac{2}{\sin x}\right) = 0\)
\(\Rightarrow \tan^{-1}\left[ \frac{\frac{2\cos x}{\sin^2 x} - \frac{2}{\sin x}}{1 + \left(\frac{2\cos x}{\sin^2 x}\right)\left(\frac{2}{\sin x}\right)} \right] = 0\)
\(\Rightarrow \frac{2\cos x\sin x - 2\sin^2x}{\sin^3x + 4\cos x} = 0\)
\(\Rightarrow 2\cos x\sin x - 2\sin^2x = 0\)
\(\Rightarrow 2\sin^2x = 2\cos x\sin x \Rightarrow \tan x = 1\)
\(\Rightarrow x = \tan^{-1}(1) = \tan^{-1}\left(\tan\frac{\pi}{4}\right) \Rightarrow x = \frac{\pi}{4}\)

Question. Prove that : \(\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right) = \frac{\pi}{4}\).
Answer: L.H.S. \(= \left[\tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{5}\right)\right] + \left[\tan^{-1}\left(\frac{1}{7}\right) + \tan^{-1}\left(\frac{1}{8}\right)\right]\)
\(= \tan^{-1}\left[ \frac{\frac{1}{3} + \frac{1}{5}}{1 - \frac{1}{3} \cdot \frac{1}{5}} \right] + \tan^{-1}\left[ \frac{\frac{1}{7} + \frac{1}{8}}{1 - \frac{1}{7} \cdot \frac{1}{8}} \right]\)
\(= \tan^{-1}\left( \frac{\frac{8}{15}}{\frac{14}{15}} \right) + \tan^{-1}\left( \frac{\frac{15}{56}}{\frac{55}{56}} \right)\)
\(= \tan^{-1}\left(\frac{8}{14}\right) + \tan^{-1}\left(\frac{15}{55}\right)\)
\(= \tan^{-1}\left(\frac{4}{7}\right) + \tan^{-1}\left(\frac{3}{11}\right)\)
\(= \tan^{-1}\left[ \frac{\frac{4}{7} + \frac{3}{11}}{1 - \frac{4}{7} \cdot \frac{3}{11}} \right]\)
\(= \tan^{-1}\left[ \frac{\frac{65}{77}}{\frac{65}{77}} \right]\)
\(= \tan^{-1}(1) = \frac{\pi}{4} = \text{R.H.S.}\)

Question. Solve the equation for \(x\) : \(\sin^{-1}x + \sin^{-1}(1-x) = \cos^{-1}x\).
Answer: \(\sin^{-1}x + \sin^{-1}(1-x) = \cos^{-1}x\)
\(\Rightarrow \sin^{-1}x + \sin^{-1}(1-x) = \frac{\pi}{2} - \sin^{-1}x\) \([\because \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}]\)
\(\Rightarrow \sin^{-1}(1-x) = \frac{\pi}{2} - 2\sin^{-1}x\)
\(\Rightarrow \sin^{-1}(1-x) = \cos^{-1}(2\sin^{-1}x)\)
\(\Rightarrow 2\sin^{-1}x = \cos^{-1}(1-x)\)
\(\Rightarrow \cos(2\sin^{-1}x) = 1-x\)
\(\Rightarrow 1 - 2\sin^2(\sin^{-1}x) = 1-x\)
\(\Rightarrow 1 - 2x^2 = 1-x\)
\(\Rightarrow 2x^2 - x = 0 \Rightarrow x(2x-1) = 0\)
\(\Rightarrow x = 0\) or \(2x - 1 = 0 \Rightarrow x = 0\) or \(x = 1/2\)

Question. If \(\cos^{-1}\frac{x}{a} + \cos^{-1}\frac{y}{b} = \alpha\), prove that \(\frac{x^2}{a^2} - 2\frac{xy}{ab}\cos\alpha + \frac{y^2}{b^2} = \sin^2\alpha\).
Answer: We have, \(\cos^{-1}\frac{x}{a} + \cos^{-1}\frac{y}{b} = \alpha\)
\(\Rightarrow \cos^{-1}\left( \frac{xy}{ab} - \sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}} \right) = \alpha\)
\(\Rightarrow \frac{xy}{ab} - \sqrt{1-\frac{x^2}{a^2}}\sqrt{1-\frac{y^2}{b^2}} = \cos\alpha\) ... (1)
Squaring on both sides, we get:
\(\left(\frac{xy}{ab} - \cos\alpha\right)^2 = \left(1-\frac{x^2}{a^2}\right)\left(1-\frac{y^2}{b^2}\right)\)
\(\Rightarrow \frac{x^2y^2}{a^2b^2} + \cos^2\alpha - 2\frac{xy}{ab}\cos\alpha = 1 - \frac{y^2}{b^2} - \frac{x^2}{a^2} + \frac{x^2y^2}{a^2b^2}\)
\(\Rightarrow \frac{x^2}{a^2} - 2\frac{xy}{ab}\cos\alpha + \frac{y^2}{b^2} = 1 - \cos^2\alpha\)
\(\Rightarrow \frac{x^2}{a^2} - 2\frac{xy}{ab}\cos\alpha + \frac{y^2}{b^2} = \sin^2\alpha\) [From (1)]

Question. Solve for \(x\) : \(\tan^{-1}\left(\frac{x-2}{x-1}\right) + \tan^{-1}\left(\frac{x+2}{x+1}\right) = \frac{\pi}{4}\).
Answer: \(\tan^{-1}\left(\frac{x-2}{x-1}\right) + \tan^{-1}\left(\frac{x+2}{x+1}\right) = \frac{\pi}{4}\)
\(\Rightarrow \tan^{-1}\left[ \frac{\left(\frac{x-2}{x-1}\right) + \left(\frac{x+2}{x+1}\right)}{1 - \left(\frac{x-2}{x-1}\right)\left(\frac{x+2}{x+1}\right)} \right] = \frac{\pi}{4}\)
\(\Rightarrow \tan^{-1}\left[ \frac{(x-2)(x+1) + (x+2)(x-1)}{(x^2-1) - (x^2-4)} \right] = \frac{\pi}{4}\)
\(\Rightarrow \tan^{-1}\left[ \frac{x^2+x-2x-2 + x^2-x+2x-2}{x^2-1-x^2+4} \right] = \frac{\pi}{4}\)
\(\Rightarrow \frac{2x^2-4}{3} = \tan\frac{\pi}{4} = 1\)
\(\Rightarrow 2x^2 - 4 = 3 \Rightarrow 2x^2 = 7\)
\(\Rightarrow x^2 = \frac{7}{2} \Rightarrow x = \pm\sqrt{\frac{7}{2}}\)

Question. Prove that : \(\cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}\), \(x \in \left(0, \frac{\pi}{4}\right)\).
Answer: L.H.S.
\(= \cot^{-1}\left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right)\)
Multiplying numerator and denominator by \(\sqrt{1+\sin x} + \sqrt{1-\sin x}\):
\(= \cot^{-1}\left[ \frac{(\sqrt{1+\sin x} + \sqrt{1-\sin x})^2}{(1+\sin x) - (1-\sin x)} \right]\)
\(= \cot^{-1}\left[ \frac{(1+\sin x) + (1-\sin x) + 2\sqrt{1-\sin^2x}}{2\sin x} \right]\)
\(= \cot^{-1}\left[ \frac{2(1+\cos x)}{2\sin x} \right] = \cot^{-1}\left( \frac{1+\cos x}{\sin x} \right)\)
\(= \cot^{-1}\left[ \frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)} \right]\)
\(= \cot^{-1}\left(\cot \frac{x}{2}\right) = \frac{x}{2} = \text{R.H.S.}\)
Hence, L.H.S. = R.H.S.

Question. If \(\sin[\cot^{-1}(x+1)] = \cos(\tan^{-1}x)\), then find \(x\).
Answer: We have, \(\sin[\cot^{-1}(x+1)] = \cos(\tan^{-1}x)\) ... (1)
Let \(\cot^{-1}(x+1) = A\) and \(\tan^{-1}x = B\)
\(\Rightarrow x + 1 = \cot A \Rightarrow \sin A = \frac{1}{\sqrt{(x+1)^2 + 1}}\)
Also, \(x = \tan B \Rightarrow \cos B = \frac{1}{\sqrt{x^2+1}}\)
Now, \(\sin A = \cos B\) [From (1)]
\(\Rightarrow \frac{1}{\sqrt{(x+1)^2 + 1}} = \frac{1}{\sqrt{x^2+1}}\)
\(\Rightarrow (x+1)^2 + 1 = x^2 + 1\)
\(\Rightarrow 1 + 2x = 0 \Rightarrow x = -\frac{1}{2}\)

Question. If \((\tan^{-1}x)^2 + (\cot^{-1}x)^2 = \frac{5\pi^2}{8}\), then find \(x\).
Answer: \((\tan^{-1}x)^2 + (\cot^{-1}x)^2 = \frac{5\pi^2}{8}\) [Given]
\(\Rightarrow (\tan^{-1}x)^2 + \left(\frac{\pi}{2} - \tan^{-1}x\right)^2 = \frac{5\pi^2}{8}\)
Putting \(\tan^{-1}x = \theta\), we get
\(\theta^2 + \left(\frac{\pi}{2} - \theta\right)^2 = \frac{5\pi^2}{8}\)
\(\Rightarrow \theta^2 + \frac{\pi^2}{4} + \theta^2 - \pi\theta = \frac{5\pi^2}{8}\)
\(\Rightarrow 2\theta^2 - \pi\theta + \left(\frac{\pi^2}{4} - \frac{5\pi^2}{8}\right) = 0\)
\(\Rightarrow 2\theta^2 - \pi\theta - \frac{3\pi^2}{8} = 0\)
\(\Rightarrow 16\theta^2 - 8\pi\theta - 3\pi^2 = 0\)
\(\Rightarrow 4\theta(4\theta - 3\pi) + \pi(4\theta - 3\pi) = 0\)
\(\Rightarrow (4\theta + \pi)(4\theta - 3\pi) = 0\)
\(\Rightarrow\) Either \(4\theta = 3\pi\) or \(4\theta = -\pi\)
\(\Rightarrow \theta = \frac{3\pi}{4}\) or \(\theta = -\frac{\pi}{4}\)
Hence, \(\tan^{-1}x = -\frac{\pi}{4}\) or \(\frac{3\pi}{4}\)
\(\Rightarrow x = -1\)

Question. Prove the following: \(\cot^{-1}\left(\frac{xy+1}{x-y}\right) + \cot^{-1}\left(\frac{yz+1}{y-z}\right) + \cot^{-1}\left(\frac{zx+1}{z-x}\right) = 0\), \((0 < xy, yz, zx < 1)\).
Answer: L.H.S.
\(= \cot^{-1}\left(\frac{xy+1}{x-y}\right) + \cot^{-1}\left(\frac{yz+1}{y-z}\right) + \cot^{-1}\left(\frac{zx+1}{z-x}\right)\)
\(= \tan^{-1}\left(\frac{x-y}{1+xy}\right) + \tan^{-1}\left(\frac{y-z}{1+yz}\right) + \tan^{-1}\left(\frac{z-x}{1+zx}\right)\) \([\because \cot^{-1}x = \tan^{-1}\frac{1}{x}]\)
\(= (\tan^{-1}x - \tan^{-1}y) + (\tan^{-1}y - \tan^{-1}z) + (\tan^{-1}z - \tan^{-1}x)\)
\(= 0 = \text{R.H.S.}\)

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Question. Solve for \( x \): \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31} \)
Answer: We have, \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31} \)
\( \Rightarrow \tan^{-1}\left( \frac{(x+1) + (x-1)}{1 - (x+1)(x-1)} \right) = \tan^{-1}\frac{8}{31} \)
for \( (x+1)(x-1) < 1 \)
\( \Rightarrow \frac{2x}{1 - (x^2 - 1)} = \frac{8}{31} \Rightarrow \frac{2x}{2 - x^2} = \frac{8}{31} \)
\( \Rightarrow 31x = 8 - 4x^2 \Rightarrow 4x^2 + 31x - 8 = 0 \)
\( \Rightarrow (4x - 1)(x + 8) = 0 \)
\( \Rightarrow 4x - 1 = 0 \) or \( x + 8 = 0 \)
\( \Rightarrow x = \frac{1}{4} \) or \( x = -8 \)
But \( x = -8 \) does not satisfy the equation.
Hence, \( x = \frac{1}{4} \) is the only solution.

Question. If \( \tan^{-1}\left(\frac{1}{1+1\cdot 2}\right) + \tan^{-1}\left(\frac{1}{1+2\cdot 3}\right) + \dots + \tan^{-1}\left(\frac{1}{1+n(n+1)}\right) = \tan^{-1}\theta \), then find the value of \( \theta \).
Answer: \( \tan^{-1}\left(\frac{1}{1+1\cdot 2}\right) + \tan^{-1}\left(\frac{1}{1+2\cdot 3}\right) + \dots + \tan^{-1}\left(\frac{1}{1+n(n+1)}\right) = \tan^{-1}\theta \)
\( \Rightarrow \tan^{-1}\left(\frac{2-1}{1+1\cdot 2}\right) + \tan^{-1}\left(\frac{3-2}{1+2\cdot 3}\right) + \dots + \tan^{-1}\left(\frac{(n+1)-n}{1+n(n+1)}\right) = \tan^{-1}\theta \)
\( \Rightarrow \tan^{-1}2 - \tan^{-1}1 + \tan^{-1}3 - \tan^{-1}2 + \dots + \tan^{-1}(n+1) - \tan^{-1}n = \tan^{-1}\theta \)
\( \Rightarrow \tan^{-1}(n+1) - \tan^{-1}1 = \tan^{-1}\theta \)
\( \Rightarrow \tan^{-1}\left(\frac{(n+1)-1}{1+(n+1)(1)}\right) = \tan^{-1}\theta \)
\( \Rightarrow \tan^{-1}\left(\frac{n}{n+2}\right) = \tan^{-1}\theta \Rightarrow \theta = \frac{n}{n+2} \)

Question. Solve for \( x \): \( \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4} \)
Answer: We have, \( \tan^{-1}2x + \tan^{-1}3x = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1}\left( \frac{2x + 3x}{1 - 2x \cdot 3x} \right) = \frac{\pi}{4} \) (for \( 2x \cdot 3x < 1 \))
\( \Rightarrow \tan^{-1}\left( \frac{5x}{1 - 6x^2} \right) = \frac{\pi}{4} \)
Therefore, \( \frac{5x}{1 - 6x^2} = \tan \frac{\pi}{4} = 1 \)
\( \Rightarrow 6x^2 + 5x - 1 = 0 \Rightarrow (6x - 1)(x + 1) = 0 \)
which gives \( x = \frac{1}{6} \) or \( x = -1 \).
Since \( x = -1 \) does not satisfy the equation as the L.H.S. of the equation becomes negative.
\( \therefore x = \frac{1}{6} \) is the only solution of the given equation.

Question. Prove that: \( \tan^{-1}\frac{63}{16} = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} \)
Answer: Consider R.H.S. \( = \sin^{-1}\frac{5}{13} + \cos^{-1}\frac{3}{5} \)
\( = \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{4}{3} \)
\( = \tan^{-1}\left( \frac{\frac{5}{12} + \frac{4}{3}}{1 - \frac{5}{12} \cdot \frac{4}{3}} \right) \)
\( = \tan^{-1}\left( \frac{\frac{15 + 48}{36}}{\frac{36 - 20}{36}} \right) = \tan^{-1}\left(\frac{63}{16}\right) = \text{L.H.S.} \)

Question. Prove that \( 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \sin^{-1}\left(\frac{31}{25\sqrt{2}}\right) \)
Answer: L.H.S. \( = 2\tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{2}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left( \frac{\frac{1}{2} + \frac{1}{2}}{1 - \frac{1}{2} \cdot \frac{1}{2}} \right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left( \frac{\frac{4}{3} + \frac{1}{7}}{1 - \frac{4}{3} \cdot \frac{1}{7}} \right) = \tan^{-1}\left( \frac{\frac{28 + 3}{21}}{\frac{21 - 4}{21}} \right) = \tan^{-1}\left(\frac{31}{17}\right) \)
Now, let \( \tan^{-1}\left(\frac{31}{17}\right) = \theta \quad \dots(1) \)
\( \Rightarrow \tan\theta = \frac{31}{17} \)
\( \therefore \sin\theta = \frac{1}{\csc\theta} = \frac{1}{\sqrt{1 + \cot^2\theta}} = \frac{1}{\sqrt{1 + \left(\frac{17}{31}\right)^2}} = \frac{31}{\sqrt{31^2 + 17^2}} = \frac{31}{\sqrt{1250}} = \frac{31}{25\sqrt{2}} \)
\( \Rightarrow \theta = \sin^{-1}\left(\frac{31}{25\sqrt{2}}\right) \quad \dots(2) \)
From (1) & (2), L.H.S \( = \sin^{-1}\left(\frac{31}{25\sqrt{2}}\right) = \text{R.H.S.} \)

Question. Solve for \( x \): \( \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\tan^{-1}x \), \( x > 0 \)
Answer: We have, \( \tan^{-1}\left(\frac{1-x}{1+x}\right) = \frac{1}{2}\tan^{-1}x \), \( (x > 0) \)
\( \Rightarrow \tan^{-1}1 - \tan^{-1}x = \frac{1}{2}\tan^{-1}x \)
\( \Rightarrow \frac{3}{2}\tan^{-1}x = \tan^{-1}1 = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1}x = \frac{\pi}{4} \times \frac{2}{3} = \frac{\pi}{6} \Rightarrow x = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} \)
\( \Rightarrow x = \frac{1}{\sqrt{3}} \)

Question. Prove that \( 2\tan^{-1}\left(\frac{1}{5}\right) + \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) + 2\tan^{-1}\left(\frac{1}{8}\right) = \frac{\pi}{4} \)
Answer: L.H.S. \( = 2\tan^{-1}\left(\frac{1}{5}\right) + \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) + 2\tan^{-1}\left(\frac{1}{8}\right) \)
\( = 2\left[ \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{8}\right) \right] + \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) \)
\( = 2 \tan^{-1}\left( \frac{\frac{1}{5} + \frac{1}{8}}{1 - \frac{1}{5}\cdot\frac{1}{8}} \right) + \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) \)
\( = 2 \tan^{-1}\left(\frac{13/40}{39/40}\right) + \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) \)
\( = 2 \tan^{-1}\left(\frac{1}{3}\right) + \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) \)
Let \( \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) = \theta \Rightarrow \sec\theta = \frac{5\sqrt{2}}{7} \Rightarrow \tan\theta = \sqrt{\sec^2\theta - 1} = \sqrt{\frac{50}{49} - 1} = \frac{1}{7} \)
\( \Rightarrow \sec^{-1}\left(\frac{5\sqrt{2}}{7}\right) = \tan^{-1}\left(\frac{1}{7}\right) \)
L.H.S. \( = 2 \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left( \frac{2 \cdot \frac{1}{3}}{1 - \left(\frac{1}{3}\right)^2} \right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left(\frac{2/3}{8/9}\right) + \tan^{-1}\left(\frac{1}{7}\right) = \tan^{-1}\left(\frac{3}{4}\right) + \tan^{-1}\left(\frac{1}{7}\right) \)
\( = \tan^{-1}\left( \frac{\frac{3}{4} + \frac{1}{7}}{1 - \frac{3}{4}\cdot\frac{1}{7}} \right) = \tan^{-1}\left(\frac{25/28}{25/28}\right) = \tan^{-1}(1) = \frac{\pi}{4} = \text{R.H.S.} \)

Question. Prove that: \( \tan^{-1}\left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x \), \( -\frac{1}{\sqrt{2}} \le x \le 1 \)
Answer: Putting \( x = \cos\theta \), we get
L.H.S. \( = \tan^{-1}\left( \frac{\sqrt{1+\cos\theta} - \sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta} + \sqrt{1-\cos\theta}} \right) \)
\( = \tan^{-1}\left( \frac{\sqrt{2\cos^2(\theta/2)} - \sqrt{2\sin^2(\theta/2)}}{\sqrt{2\cos^2(\theta/2)} + \sqrt{2\sin^2(\theta/2)}} \right) \)
\( = \tan^{-1}\left( \frac{\cos(\theta/2) - \sin(\theta/2)}{\cos(\theta/2) + \sin(\theta/2)} \right) \)
[Dividing numerator and denominator by \( \cos(\theta/2) \)]
\( = \tan^{-1}\left( \frac{1 - \tan(\theta/2)}{1 + \tan(\theta/2)} \right) \)
\( = \tan^{-1}\left[ \tan\left(\frac{\pi}{4} - \frac{\theta}{2}\right) \right] = \frac{\pi}{4} - \frac{\theta}{2} \)
\( = \frac{\pi}{4} - \frac{1}{2}\cos^{-1}x = \text{R.H.S.} \)

Question. If \( \tan^{-1}\left(\frac{x-2}{x-4}\right) + \tan^{-1}\left(\frac{x+2}{x+4}\right) = \frac{\pi}{4} \), find the value of \( x \).
Answer: We have, \( \tan^{-1}\left(\frac{x-2}{x-4}\right) + \tan^{-1}\left(\frac{x+2}{x+4}\right) = \frac{\pi}{4} \)
\( \Rightarrow \tan^{-1}\left( \frac{\frac{x-2}{x-4} + \frac{x+2}{x+4}}{1 - \frac{x-2}{x-4}\cdot\frac{x+2}{x+4}} \right) = \frac{\pi}{4} \)
\( \Rightarrow \frac{(x-2)(x+4) + (x+2)(x-4)}{(x-4)(x+4) - (x-2)(x+2)} = \tan\frac{\pi}{4} = 1 \)
\( \Rightarrow \frac{(x^2 + 2x - 8) + (x^2 - 2x - 8)}{(x^2 - 16) - (x^2 - 4)} = 1 \)
\( \Rightarrow \frac{2x^2 - 16}{-12} = 1 \)
\( \Rightarrow 2x^2 - 16 = -12 \Rightarrow 2x^2 = 4 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2} \)