If $y= an ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x} ight)$, then $\frac{d y}{d x}=$

Question

If $y=	an ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}ight)$, then $\frac{d y}{d x}=$

Aicharya · Accepted Answer

Given $y=	an ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}ight)=	an ^{-1}\left(\frac{2 \sin x \cos x}{2 \cos ^{2} x}ight)=	an ^{-1}(	an x)$ $	herefore \quad y=x \Rightarrow \frac{d y}{d x}=1$

If $y=\tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right)$, then $\frac{d y}{d x}=$

Solution