We are given $f'(x)=\tan^{-1}(\sec x+\tan x), \qquad -\frac{\pi}{2}<x<\frac{\pi}{2},$ and $f(0)=0.$

We need to find $f(1)$.

First simplify $f'(x)$.

Use the identity $\sec x+\tan x=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right),$ valid for $-\frac{\pi}{2}<x<\frac{\pi}{2}$.

Hence, $f'(x)=\tan^{-1}\left(\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right).$

Since $-\frac{\pi}{2}<x<\frac{\pi}{2}$, we have $0<\frac{\pi}{4}+\frac{x}{2}<\frac{\pi}{2}.$ This lies in the principal range of $\tan^{-1}$, so $\tan^{-1}\left(\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right)=\frac{\pi}{4}+\frac{x}{2}.$

Therefore, $f'(x)=\frac{\pi}{4}+\frac{x}{2}.$

Integrate:

Use the condition $f(0)=0$: $0=\frac{\pi\cdot 0}{4}+\frac{0^2}{4}+C \implies C=0.$

So, $f(x)=\frac{\pi x}{4}+\frac{x^2}{4}.$

Now compute $f(1)$: $f(1)=\frac{\pi}{4}+\frac{1}{4}=\frac{\pi+1}{4}.$

Compare with options: $\frac{\pi+1}{4}$ corresponds to Option C.

Comparison with stored correct answer: Stored correct answer is C, which matches our derived answer.