JEE Main 2021 (Online) 27th August Evening ShiftQuestion 1
If $y(x)=\cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$, $x\in\left(\frac{\pi}{2},\pi\right)$, then $\frac{dy}{dx}$ at $x=\frac{5\pi}{6}$ is :
Inverse Trigonometric Functions
Solution
For $x\in(\frac\pi2,\pi)$, $\frac x2\in(\frac\pi4,\frac\pi2)$: $\sqrt{1+\sin x}=\sin\frac x2+\cos\frac x2$, $\sqrt{1-\sin x}=\sin\frac x2-\cos\frac x2$. Ratio $=\frac{2\sin(x/2)}{2\cos(x/2)}=\tan\frac x2$. $y=\cot^{-1}(\tan\frac x2)=\frac\pi2-\frac x2$ (in $(0,\pi)$). So $\frac{dy}{dx}=-\frac12$. Option (B).