Choose the correct alternative in the following:

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

Dividing Numerator and denominator by $\cos x$ we get,

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

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

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

$y=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+x\right)\right] \because \frac{\tan a+\tan b}{1-\tan a \cdot \tan b}=\tan (a+b)$

$y=\frac{\pi}{4}+x$

Differentiating w.r.t $\mathrm{x}$ we get,

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