Solve

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

$\Rightarrow \tan ^{-1} 1-\tan ^{-1} x=\frac{1}{2} \tan ^{-1} x \quad\left[\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}\right]$

$\Rightarrow \frac{\pi}{4}=\frac{3}{2} \tan ^{-1} x$

$\Rightarrow \tan ^{-1} x=\frac{\pi}{6}$

$\Rightarrow x=\tan \frac{\pi}{6}$

$\therefore x=\frac{1}{\sqrt{3}}$