Question

If $ \tanh^{-1} x = \coth^{-1} y = \log \sqrt{5} $, then find $ \tan^{-1}(xy) = ? $

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{3} \)
• C. \( \frac{\pi}{6} \)
• D. \( \frac{3\pi}{4} \)

Hint:

Use exponential definitions for \( \tanh^{-1} x \) and \( \coth^{-1} x \), and simplify using hyperbolic identities.

Answer 1

The Correct Option is A

Given: \[ \tanh^{-1} x = \log \sqrt{5} \Rightarrow x = \tanh(\log \sqrt{5}) \Rightarrow x = \frac{e^{\log \sqrt{5}} - e^{-\log \sqrt{5}}}{e^{\log \sqrt{5}} + e^{-\log \sqrt{5}}} = \frac{\sqrt{5} - \frac{1}{\sqrt{5}}}{\sqrt{5} + \frac{1}{\sqrt{5}}} = \frac{5 - 1}{5 + 1} = \frac{2}{3} \] Similarly, \[ \coth^{-1} y = \log \sqrt{5} \Rightarrow y = \coth(\log \sqrt{5}) = \frac{\sqrt{5} + \frac{1}{\sqrt{5}}}{\sqrt{5} - \frac{1}{\sqrt{5}}} = \frac{5 + 1}{5 - 1} = \frac{6}{4} = \frac{3}{2} \] So: \[ xy = \frac{2}{3} \cdot \frac{3}{2} = 1 \Rightarrow \tan^{-1}(xy) = \tan^{-1}(1) = \frac{\pi}{4} \]