Question

Prove that:
${\tan }^{-1}\left(\frac{1+x}{1-x}\right)=\frac{\pi }{4}+{\tan }^{-1}x,x<1$

Answer 1

$LHS={\tan }^{-1}\left(\frac{1+x}{1-x}\right)$
$={\tan }^{-1}\left(\frac{1+\tan \theta }{1-\tan \theta }\right)$ Put $x=\tan \theta$
$={\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\cdot \tan \theta }\right)$ $[\because \tan \frac{\pi }{4}=1]$
$={\tan }^{-1}\left(\tan (\frac{\pi }{4}+\theta )\right)$
$=\frac{\pi }{4}+\theta$
$=\frac{\pi }{4}+{\tan }^{-1}x$ $[\because \tan \theta =x\Rightarrow \theta ={\tan }^{-1}x]$
$=RHS$
$\therefore LHS=RHS$ ......proved