$\begin{aligned} f^{\prime}(x) & =\tan ^{-1}(\sec x+\tan x) \\ & =\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) \\ & =\tan ^{-1}\left[\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}{\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}}\right] \\ & =\tan ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right) \\ & =\tan ^{-1}\left[\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) \\ & =\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]\end{aligned}$
$\begin{aligned} \therefore \quad \mathrm{f}^{\prime}(x) & =\frac{\pi}{4}+\frac{x}{2} \\ \Rightarrow \mathrm{f}(x) & =\int\left(\frac{\pi}{4}+\frac{x}{2}\right) \mathrm{d} x \\ & =\frac{\pi x}{4}+\frac{1}{2} \cdot \frac{x^2}{2}+\mathrm{c}\end{aligned}$
$\begin{array}{ll}\therefore \quad f(0)=c & \\ \Rightarrow c=0 & \ldots[\because f(0)=0 \text { (given) }]\end{array}$
$\begin{array}{r}\therefore \quad \mathrm{f}(x)=\frac{\pi x}{4}+\frac{x^2}{4} \\ \Rightarrow \mathrm{f}(1)=\frac{\pi+1}{4}\end{array}$