Question

Find the value of x, If

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

Answer 1

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

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

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

Replacing $x$ by $\frac{1-x}{1+x}$

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

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

$={\tan }^{-1}[\frac{2(1-x)}{(1+x)}\times \frac{(1+x)}{{(1+x)}^2-{(1-x)}^2}]$

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

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

$={\tan }^{-1}\left[\frac{2(1-x^2)}{4x}\right]$

$={\tan }^{-1}\left[\frac{1-x^2}{2x}\right]$

$\therefore 2{\tan }^{-1}\frac{1-x}{1+x}={\tan }^{-1}\left[\frac{1-x^2}{2x}\right]$

From $\left(1\right)$

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

Putting value

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

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

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

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

$\Rightarrow 3x^2=1$

$\Rightarrow x^2=\frac{1}{3}$

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

$x=\frac{-1}{\sqrt{3}}$ is not possible because

Given that $x>0$

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