The differential coefficient of $\tan ^{1-}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$ w.r.t. x is :

1

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

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

$\Rightarrow \frac{d y}{d x}=1$

Hence (3) is correct answer.