Question

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

Answer 1

Given:
${\tan }^{-1}\frac{1-x}{1+x}=\frac{1}{2}{\tan }^{-1}x$
$2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\tan }^{-1}x$
$\{\text{Using}2{\tan }^{-1}x={\tan }^{-1}\left(\frac{2x}{1-x^2}\right)\}$
$\Rightarrow {\tan }^{-1}\left[\frac{2\left(\frac{1-x}{1+x}\right)}{1-{\left(\frac{1-x}{1+x}\right)}^2}\right]={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left[\frac{\frac{2(1-x)}{(1+x)}}{\frac{(1+x{)}^2-(1-x{)}^2}{(1+x{)}^2}}\right]={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}[\frac{2(1-x)}{(1+x)}\times \frac{(1+x^2)}{(1+x{)}^2-(1-x{)}^2}]={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left[\frac{2(1-x)(1+x)}{(1+x{)}^2-(1-x{)}^2}\right]={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left[\frac{2(1-x^2)}{(1+x+1-x)(1+x-1+x)}\right]={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left[\frac{2(1-x^2)}{(1+1+x-x)(x+x-1+1)}\right]={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left[\frac{2(1-x^2)}{\left(2\right)\left(2x\right)}\right]={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left[\frac{1-x^2}{2x}\right]={\tan }^{-1}x$
$\Rightarrow \frac{1-x^2}{2x}=x$
$\Rightarrow 1-x^2=2x^2$
$\Rightarrow 1=x^2+2x^2$
$\Rightarrow 3x^2=1$
$\Rightarrow x^2=\frac{1}{3}$
$\Rightarrow x=\pm \frac{1}{\sqrt{3}}$
$x=\frac{-1}{\sqrt{3}}$ is not possible because it is given that $x>0$.
Hence, the value of $x$ is $\frac{1}{\sqrt{3}}$.