If $tan^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}tan^{-1}x, x> 0),$ then value of x is :

$\frac{1}{\sqrt{3}}$

The correct answer is option (3) → $\frac{1}{\sqrt{3}}$

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

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

$⇒\tan^{-1}(1)-\tan^{-1}(x)=\frac{\tan^{-1}x}{2}$

so $\frac{π}{4}=\frac{3}{2}\tan^{-1}x$

$\tan^{-1}x=\frac{π}{6}⇒x=\frac{1}{\sqrt{3}}$