Question

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

• A. $1/2$
• B. $\pi / 4$
• C. $0$
• D. $1$

Answer 1

Answer: (D)

We have,
$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) $
$=\tan ^{-1}\left(\frac{\tan \pi / 4+\tan x}{1-\tan \pi / 4 \tan x}\right)$
$=\tan ^{-1} \tan \left(\frac{\pi}{4}+x\right)=\frac{\pi}{4}+x $
$\frac{d y}{d x} =1$