Choose the correct alternative in the following:

Question:

Choose the correct alternative in the following:

If $y=\tan ^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$, then $\frac{d y}{d x}$ is equal to

A. $\frac{1}{2}$

B. 0

C. 1

D. none of these

Solution:

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

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,

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

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