Question

If $2{\tan }^{-1}\left(\frac{1+x}{1-x}\right)=5{\tan }^{-1}x,$ then value of $x$ is

• A. $\tan \left(\frac{1}{\sqrt{3}}\right)$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{\pi }{4}$
• D. $\sqrt{3}$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{3}}$

Given, $2{\tan }^{-1}\left(\frac{1+x}{1-x}\right)=5{\tan }^{-1}x,$

Put $x=\tan \theta$

$2{\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right)=5{\tan }^{-1}\left(\tan \theta \right)$

$\Rightarrow 2{\tan }^{-1}\left(\tan (\frac{\pi }{4}+\theta )\right)=5\theta$

$\Rightarrow \frac{\pi }{2}+2\theta =5\theta$

$\Rightarrow \theta =\frac{\pi }{6}$

$\therefore x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$