SOLUTION : Given Equation is $\tan ^{-1}\left(\frac{1-x}{1+x}\right)-\frac{1}{2} \tan ^{-1} x=0 ; x>0$
$2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\tan ^{-1} x {\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^2}\right]} \\$
$\tan ^{-1}\left[\frac{2\left(\frac{1-x}{1+x}\right)}{1-\left(\frac{1-x}{1+x}\right)^2}\right]=\tan ^{-1} x \Rightarrow \quad \frac{2\left(\frac{1-x}{1+x}\right)}{1-\frac{(1-x)^2}{(1+x)^2}}=x=\frac{2\left(\frac{1-x}{1+x}\right)}{\frac{(1+x)^2-(1-x)^2}{(1+x)^2}}=x$
$\quad \frac{2(1-x)}{(1+x)} \times \frac{(1+x)^2}{1+x^2+2 x-1-x^2+2 x}=x \Rightarrow \frac{2(1-x)(1+x)}{4 x}=x \\$
$2 x^2=(1-x)(1+x) \quad \Rightarrow \quad 2 x^2=1-x^2 \quad \Rightarrow \quad x^2=\frac{1}{3} \\$
$\therefore \quad x= \pm \frac{1}{\sqrt{3}} \quad \Rightarrow \quad x=\frac{1}{\sqrt{3}} \\$
