Question

If $y=ta{n}^{-1}\left(\frac{sinx+cosx}{cosx-sinx}\right)$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

The correct option is A 1

$y={\tan }^{-1}\left[\frac{\sin x+\cos x}{\cos x-\sin x}\right]$
Let $z=\frac{\sin x+\cos x}{\cos x-\sin x}$ and $y={\tan }^{-1}z$
Divide numerator and denominator by cosx
$z=\frac{\tan x+1}{1-\tan x}$
$z=\frac{\tan x+\tan \pi /4}{1-\tan x\cdot \tan \pi /4}$
$z=\tan (x+\frac{\pi }{4})$
$\therefore y={\tan }^{-1}\left[\tan (x+\frac{\pi }{4})\right]$
$y=x+\frac{\pi }{4}$
Differentiate both sides w.r.t. x
$\frac{dy}{dx}=1$