JEE Main 2019 (Online) 12th April Evening ShiftQuestion 1
The derivative of $\tan^{-1}\left(\frac{\sin x-\cos x}{\sin x+\cos x}\right)$ with respect to $\frac x2$, where $x\in\left(0,\frac\pi2\right)$, is
Inverse Trigonometric Functions
Solution
$\frac{\sin x-\cos x}{\sin x+\cos x}=\tan\left(x-\frac\pi4\right)$ with $x-\frac\pi4\in(-\frac\pi4,\frac\pi4)$, so the function is $x-\frac\pi4$; its derivative w.r.t. $\frac x2$ is $2$.