If $y= an ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$, then $\frac{d y}{d x}$ is

Question

If $y=	an ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$, then $\frac{d y}{d x}$ is

Aicharya · Accepted Answer

$\begin{aligned} & y=	an ^{-1} \sqrt{\frac{1+\cos x}{1- \cos x}}=	an ^{-1} \sqrt{\frac{2 \cos ^2 \frac{x}{2}}{2 \sin ^2 \frac{x}{2}}}=	an ^{-1}\left(\cot \frac{x}{2}ight) \ & \Rightarrow y=	an ^{-1} 	an \left(\frac{\pi}{2}-\frac{x}{2}ight)=\frac{\pi}{2}-\frac{x}{2}\end{aligned}$ Differentiating $\frac{\mathrm{dy}}{\mathrm{dx}}=0-\frac{1}{2}=\frac{-1}{2}$

If $y=\tan ^{-1} \sqrt{\frac{1+\cos x}{1-\cos x}}$, then $\frac{d y}{d x}$ is

Solution