If $y= an ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}} ight]$, then $\frac{d y}{d x}=$

Question

If $y=	an ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}ight]$, then $\frac{d y}{d x}=$

Aicharya · Accepted Answer

Given $y=	an ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}ight]$ $=	an ^{-1} \sqrt{\frac{2 \cos ^{2} \frac{x}{4}}{2 \sin ^{2} \frac{x}{4}}}=	an ^{-1}\left(\cot \frac{x}{4}ight)=	an ^{-1}\left[	an \left(\frac{\pi}{2}-\frac{x}{4}ight)ight]$ $	herefore y=-\frac{x}{4} \Rightarrow \frac{d y}{d x}=\frac{-1}{4}$

If $y=\tan ^{-1}\left[\sqrt{\frac{1+\cos \frac{x}{2}}{1-\cos \frac{x}{2}}}\right]$, then $\frac{d y}{d x}=$

Solution