Question

Solve ${\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}x,(0<x<1)$

Answer 1

${\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}x\Rightarrow 2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left(\frac{2\left(\frac{1-x}{1+x}\right)}{1-{\left(\frac{1-x}{1+x}\right)}^2}\right)={\tan }^{-1}x\Rightarrow {\tan }^{-1}\left(\frac{1-x^2}{2x}\right)={\tan }^{-1}x$
$\Rightarrow \frac{1-x^2}{2x}=x\Rightarrow 1-x^2=2x^2\Rightarrow x=\pm \frac{1}{\sqrt{3}}$
Neglecting $x=-\frac{1}{\sqrt{3}}$
We get $x=\frac{1}{\sqrt{3}}$