Question

Differentiate, ${\tan }^{-1}\left[\frac{\cos x+\sin x}{\cos x-\sin x}\right]$ with respect to $x$.

Answer 1

Suppose that
$y={\tan }^{-1}\left[\frac{\cos x+\sin x}{\cos x-\sin x}\right]$
$y={\tan }^{-1}\left[\frac{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}\right]$
$\Rightarrow y={\tan }^{-1}\left[\frac{1+\tan x}{1-\tan x}\right]$
$\Rightarrow y={\tan }^{-1}\left[\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}.\tan x}\right]$
$\Rightarrow y={\tan }^{-1}\left[\tan (\frac{\pi }{4}+x)\right]$
$\Rightarrow y=\frac{\pi }{4}+x$
$\therefore \frac{dy}{dx}=\frac{d}{dx}\frac{\pi }{4}+\frac{d}{dx}.x$
$\Rightarrow \frac{dy}{dx}=0+1$
$\Rightarrow \frac{dy}{dx}=1$.