Question

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

Answer 1

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

$\Rightarrow 2{\tan }^{-1}\frac{1-x}{1+x}-{\tan }^{-1}x=0$ multiply both sides by $2$

We know that $2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)=0$

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

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

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

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

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

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

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

$\Rightarrow 3x^2=1$

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

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

Given that $x>0$ the value $x$ should be positive.

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