Show that the derivative of $\displaystyle \tan ^{-1}(\sec \mathrm{x}+\tan \mathrm{x}),\left[-\frac{\pi}{2}<\mathrm{x}<\frac{\pi}{2}\right]$ with respect to x is equal to $\displaystyle \frac{1}{2}$.

$\displaystyle \tan ^{-1}(\sec \mathrm{x}+\tan \mathrm{x})=\tan ^{-1}\left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}\right)=\tan ^{-1}\left(\frac{1+\cos \left(\dfrac{\pi}{2}-\mathrm{x}\right)}{\sin \left(\dfrac{\pi}{2}-\mathrm{x}\right)}\right)$ \[\begin{aligned} & =\tan ^{-1}\left(\frac{2 \cos ^{2}\left(\dfrac{\pi}{4}-\dfrac{\mathrm{x}}{2}\right)}{2 \sin \left(\dfrac{\pi}{4}-\dfrac{\mathrm{x}}{2}\right) \cos \left(\dfrac{\pi}{4}-\dfrac{\mathrm{x}}{2}\right)}\right) \\ & =\tan ^{-1}\left(\cot \left(\frac{\pi}{4}-\frac{\mathrm{x}}{2}\right)\right) \\ & =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{\mathrm{x}}{2}\right)\right)=\frac{\pi}{4}+\frac{\mathrm{x}}{2} \\ & \therefore\left(\tan ^{-1}(\sec \mathrm{x}+\tan \mathrm{x})\right)^{\prime}=\frac{1}{2} \end{aligned} \]