Question

Find the value of x, if
${\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}x,x>0.$

• A. 1
• B. $\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. No solution

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}$

${\tan }^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}{\tan }^{-1}x\Rightarrow 2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\tan }^{-1}x\mathrm{We}\mathrm{know}\mathrm{that}{\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+x.y}\right)\mathrm{So},\mathrm{we}\mathrm{can}\mathrm{write}{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\tan }^{-1}1-{\tan }^{-1}x\mathrm{Therefore},2[{\tan }^{-1}1-{\tan }^{-1}x]={\tan }^{-1}x\Rightarrow 3{\tan }^{-1}x=2\left(\frac{\pi }{4}\right)\Rightarrow {\tan }^{-1}x=\frac{\pi }{6}\Rightarrow x=\tan \frac{\pi }{6}\Rightarrow x=\frac{1}{\sqrt{3}}$