Question

$\frac{d}{dx}\left({\tan }^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}\right)$

• A. $-1$
• B. $-2$
• C. $1$
• D. $x$

Answer 1

The correct option is A $-1$

Let $y={\tan }^{-1}\frac{\cos x-\sin x}{\cos x+\sin x}={\tan }^{-1}\frac{1-\tan x}{1+\tan x}$
$\Rightarrow y={\tan }^{-1}\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\tan x}={\tan }^{-1}\tan (\frac{\pi }{4}-x)=\frac{\pi }{4}-x$
$\therefore \frac{dy}{dx}=-1$