Question

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

Answer 1

Given,

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

or, $y={\tan }^{-1}\frac{1+\tan x}{1-\tan x}$ [ Dividing the numerator and the denominator by $\cos x$]

or, $y={\tan }^{-1}\frac{\tan \frac{\pi }{4}+\tan x}{1-\tan \frac{\pi }{4}\tan x}$

or, $y={\tan }^{-1}\tan (\frac{\pi }{4}-x)$ [ Since $\tan \frac{\pi }{4}=1$]

or, $y=\frac{\pi }{4}+x$

So $\frac{dy}{dx}=1$.