Answer
$f(t)=\tan t+\sec t-2-\sqrt{2}$
Work Step by Step
$f'(t)=\sec^{2}t+\sec t\tan t$
Using the antiderivatives table,
$f(t)=\tan t+\sec t+C$
Given that $f(\pi/4)=-1$, we find $C$:
$-1=\tan(\pi/4)+\sec(\pi/4)+C$
$-1=1+\sqrt{2}+C$
$C=-2-\sqrt{2}$
$f(t)=\tan t+\sec t-2-\sqrt{2}$