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

Given
\(\begin{aligned} y &=\tan ^{-1}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) \\ &=\tan ^{-1}\left[\frac{\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}\right)+2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}\right) \\ &=\tan ^{-1}\left\{\frac{x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\right\} \\ &=\tan ^{-1}\left\{\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right\}=\tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) \\ &=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right) \end{aligned}\)
\(\begin{aligned} \mathrm{y} &=\frac{\pi}{4}+\frac{\mathrm{x}}{2} \\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}} &=0+\frac{1}{2}=\frac{1}{2} \end{aligned}\)