Question

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

• A. $1$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. None of these

Answer 1

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

${\tan }^{-1}\frac{1-x}{1+x}=\frac{1}{2}{\tan }^{-1}x\Rightarrow 2{\tan }^{-1}\left(\frac{1-x}{1+x}\right)={\tan }^{-1}x$
$\Rightarrow {\tan }^{-1}\left(\frac{2\left(\frac{1-x}{1+x}\right)}{1-{\left(\frac{1-x}{1+x}\right)}^2}\right)={\tan }^{-1}x\Rightarrow \frac{2(1-x)(1+x)}{{(1+x)}^2-{(1-x)}^2}=x$
$\Rightarrow \frac{2(1-x^2)}{4x}=x\Rightarrow 2-{2x}^2={4x}^2$
$\Rightarrow 2={6x}^2\Rightarrow x=\frac{1}{\sqrt{3}}$