Question

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

• A. $\frac{1}{2}$
• B. $\frac{\pi }{4}$
• C. $0$
• D. $1$

Answer 1

The correct option is D $1$

$y={\tan }^{-1}\left(\frac{\sin x+\cos x}{\cos x-\sin x}\right)\text{Divide and multiply the denominator and numerator by}\cos \theta \Rightarrow y={\tan }^{-1}\left(\frac{\tan x+1}{1-\tan x}\right)={\tan }^{-1}\left(\tan (\frac{\pi }{4}+x)\right)\Rightarrow y=\frac{\pi }{4}+x\therefore \frac{dy}{dx}=1$