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

$-1$

The correct answer is Option (2) → $-1$ ##

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

$= \tan^{-1} \left[ \frac{\cos x \left( 1 - \frac{\sin x}{\cos x} \right)}{\cos x \left( 1 + \frac{\sin x}{\cos x} \right)} \right]$

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

$=\tan^{-1} \left[\tan\left(\frac{\pi}{4}-x\right)\right]$

$=\frac{\pi}{4}-x$

$⇒\frac{dy}{dx}=-1$