Question

${\tan }^{-1}\frac{1-x}{1+x}=\frac{1}{2}{\tan }^{-1}x=0,x>0$.

Answer 1

Let ${\tan }^{-1}x=A\Rightarrow \tan A=x$

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

$\Rightarrow {\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}-\tan A}{1+\tan \frac{\pi }{4}}\right)$

$\Rightarrow {\tan }^{-1}\left(\tan (\frac{\pi }{4}-A(\right)$

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

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

$\Rightarrow x=\tan (\frac{\pi }{2}-2A)$

$=\cot 2A$

$=\frac{1}{\tan 2A}$

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

$\Rightarrow 2x^2=1-x^2$

$\Rightarrow 3x^2=1$

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

Hence, solved.