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,
$y={\tan }^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)$
$={\tan }^{-1}\left(\frac{1+\tan x}{1-\tan x}\right)$
$={\tan }^{-1}\left(\frac{\tan \pi /4+\tan x}{1-\tan \pi /4\tan x}\right)$
$={\tan }^{-1}\tan \left(\frac{\pi }{4}+x\right)=\frac{\pi }{4}+x$
$\frac{dy}{dx}=1$