Question

If $y={\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$, then $\frac{dy}{dx}=$

• A. $-1$
• B. $\sin 2\pi$
• C. $\cos 2x$
• D. Zero

Answer 1

The correct option is A $-1$

We have, $y={\tan }^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)$
$y={\tan }^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)$ $\left[\text{dividing numerator and denominator by cosx}\right]$
$y={\tan }^{-1}\left(\frac{\tan \frac{\pi }{4}-\tan x}{1+\tan \frac{\pi }{4}\cdot \tan x}\right)$
$y={\tan }^{-1}\left(\tan (\frac{\pi }{4}-x)\right)$
$y=\frac{\pi }{4}-x$
$\therefore \frac{dy}{dx}=-1$