If $y= an ^{-1}(\sec x- an x)$, then $\frac{d y}{d x}=$

Question

If $y=	an ^{-1}(\sec x-	an x)$, then $\frac{d y}{d x}=$

Aicharya · Accepted Answer

$\begin{aligned} & y=	an ^{-1}(\sec x-	an x)=	an ^{-1}\left(\frac{1-\sin x}{\cos x}ight)=	an ^{-1}\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}ight) \ & \Rightarrow y=	an ^{-1}\left(\frac{1-	an \frac{x}{2}}{1+	an \frac{x}{2}}ight)=	an ^{-1}\left(	an \left(\frac{\pi}{4}-\frac{x}{2}ight)ight)=\frac{\pi}{4}-\frac{x}{2} \ & \Rightarrow \frac{d y}{d x}=\frac{-1}{2}\end{aligned}$

If $y=\tan ^{-1}(\sec x-\tan x)$, then $\frac{d y}{d x}=$

Solution