Question

Write the values of $x$ for which $2{\tan }^{-1}x={\cos }^{-1}\frac{1-x^2}{1+x^2}$, holds.

Answer 1

Given : $2{\tan }^{-1}x={\cos }^{-1}\frac{1-x^2}{1+x^2}$

Substitute $x=\tan \theta$

$⟹2{\tan }^{-1}x=2{\tan }^{-1}\left(\tan \theta \right)2\theta$

${\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$={\cos }^{-1}\left(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\right)$

$={\cos }^{-1}\left(\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta -{\sin }^2\theta }\right)={\cos }^{-1}\left(\cos 2\theta \right)=2\theta =\text{LHS}$

$\therefore \text{LHS=RHS}$ holds for $x=\tan \theta$

$\because \tan \theta \in (0,\infty )$

$⟹x\in (0,\infty )$