Given, $0<x<1$

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

Put $x=\tan \left(\theta \right) ; 0<\theta <\frac{\mathrm{\pi }}{4}$

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

$\Rightarrow 2{\tan }^{-1}\left(\tan \left(\frac{\pi }{4}-\theta \right)\right)={\cos }^{-1}\left(\cos 2\theta \right)$

$2\left(\frac{\pi }{4}-\theta \right)=2\theta \Rightarrow \theta =\frac{\pi }{8}$

Put $\theta ={\tan }^{-1}x$ we get,

$\mathrm{x}=\tan \frac{\pi }{8}\therefore \mathrm{x}=\sqrt{2}-1\simeq 0.414<\frac{1}{2}$