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

We have, ${\tan ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = \frac{1}{2}{\tan ^{ - 1}}x,\;\;(x > 0)$

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

$\Rightarrow$ ${\tan ^{ - 1}}x = \frac{\pi }{4} \times \frac{2}{3} = \frac{\pi }{6} \Rightarrow \,x = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }} \Rightarrow x = \frac{1}{{\sqrt 3 }}$