Question

Solve the following equation for x:

$ta{n}^{-1}\frac{1-x}{1+x}=\frac{1}{2}ta{n}^{-1}x,(x>0)$.

Answer 1

Given,
$ta{n}^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}ta{n}^{-1}xor2ta{n}^{-1}\left(\frac{1-x}{1+x}\right)=ta{n}^{-1}x\Rightarrow ta{n}^{-1}\left[\frac{2\left(\frac{1-x}{1+x}\right)}{1-{\left(\frac{1-x}{1+x}\right)}^2}\right]=ta{n}^{-1}x[\because 2ta{n}^{-1}y=ta{n}^{-1}\left(\frac{2y}{1-y^2}\right)]\Rightarrow ta{n}^{-1}\left[\frac{\frac{2(1-x)}{(1+x)}}{\frac{(1+x{)}^2-(1-x{)}^2}{(1+x{)}^2}}\right]=ta{n}^{-1}x\Rightarrow \left[\frac{2(1-x)(2+x)}{(1+x{)}^2-(1-x{)}^2}\right]=ta{n}^{-1}x\Rightarrow ta{n}^{-1}\left[\frac{2(1-x{)}^2}{1^2+x^2+2x-1^2-x^2+2x}\right]=ta{n}^{-1}x\Rightarrow ta{n}^{-1}\left[\frac{2(1-x^2)}{4x}\right]=ta{n}^{-1}x\Rightarrow ta{n}^{-1}\left(\frac{2-x^2}{2x}\right)=ta{n}^{-1}x\Rightarrow \frac{1-x^2}{2x}=x\Rightarrow 1-x^2=2x^2\Rightarrow 1=3x^2\Rightarrow x^2=\frac{1}{3}\Rightarrow x=\pm \frac{1}{\sqrt{3}}$
[$\because x>0$ given, so we do not take $x=-\frac{1}{\sqrt{3}}$]
$\Rightarrow x=\frac{1}{\sqrt{3}}$