Question

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

Answer 1

We have,

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

Put, $x=\tan \theta$ then, $\theta ={\tan }^{-1}x$

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

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

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

${\tan }^{-1}\tan (\frac{\pi }{4}-\theta )=\frac{1}{2}\theta$

$(\frac{\pi }{4}-\theta )=\frac{1}{2}\theta$

$\frac{\pi }{4}-\theta =\frac{1}{2}\theta$

$\frac{\pi }{4}=\frac{1}{2}\theta +\theta$

$\frac{\pi }{4}=\frac{3}{2}\theta$

$3\theta =\frac{\pi }{2}$

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

Now, put $\theta ={\tan }^{-1}x$

${\tan }^{-1}x=\frac{\pi }{6}$

$x=\tan \frac{\pi }{6}$

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

Hence, this is the answer.