Question

If $y={\tan }^{-1}\frac{\cos x+\sin x}{cosx+\sin x}$ then find $\frac{dy}{dx}$

Answer 1

We have,

$y={\tan }^{-1}\frac{\cos x+\sin x}{\cos x+\sin x}$

$y={\tan }^{-1}1$

$y={\tan }^{-1}\tan \frac{\pi }{4}$

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

On differentiation with respect to x and we get,

$\frac{dy}{dx}=\frac{d}{dx}\left(\frac{\pi }{4}\right)$

$\frac{dy}{dx}=0$

Hence, this is the answer.