Question

If \( 2 \tan^{-1} (\cos x) = \tan^{-1} (2 \csc x) \), then find the value of \( x \).

Hint:

Simplify trigonometric equations using identities; verify solutions to ensure consistency.

Answer 1

Let \( \theta = \tan^{-1} (\cos x) \), so \( 2\theta = \tan^{-1} (2 \csc x) \). 
\[ \tan \theta = \cos x, \tan 2\theta = 2 \csc x. \] Using double angle formula: \( \tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta} \). 
\[ \frac{2 \tan \theta}{1 - \tan^2 \theta} = 2 \csc x. \] Since \( \tan \theta = \cos x \), and \( \csc x = \frac{1}{\sin x} \), we have: 
\[ \frac{2 \cos x}{1 - \cos^2 x} = \frac{2}{\sin x}. \] Since \( \sin^2 x = 1 - \cos^2 x \): 
\[ \frac{2 \cos x}{\sin^2 x} = \frac{2}{\sin x} \Rightarrow \frac{\cos x}{\sin^2 x} = \frac{1}{\sin x} \Rightarrow \cos x = \sin x \Rightarrow \tan x = 1. \] \[ x = \frac{\pi}{4} + k\pi, k \in \mathbb{Z}. \] Principal solution: \( x = \frac{\pi}{4} \). 
Verify: Left: \( 2 \tan^{-1} (\cos \frac{\pi}{4}) = 2 \tan^{-1} \left( \frac{\sqrt{2}}{2} \right) = 2 \cdot \frac{\pi}{8} = \frac{\pi}{4} \). 
Right: \( \tan^{-1} \left( 2 \csc \frac{\pi}{4} \right) = \tan^{-1} \left( 2 \cdot \sqrt{2} \right) \approx \tan^{-1} (2.828) \approx \frac{\pi}{4} \). Matches. 
Answer: \( x = \frac{\pi}{4} \).