Answer
$$
f^{\prime}(t)=\sec t(\sec t+\tan t),\quad -\frac{\pi}{2} \lt t \lt \frac{\pi}{2}, \quad f(\frac{\pi}{4})=-1
$$
The required function is
$$
f(t)=\tan t+\sec t+-2-\sqrt {2}
$$
Work Step by Step
$$
f^{\prime}(t)=\sec t(\sec t+\tan t),\quad -\frac{\pi}{2} \lt t \lt \frac{\pi}{2}, \quad f(\frac{\pi}{4})=-1
$$
can be written as the following:
$$
f^{\prime}(t)=\sec t(\sec t+\tan t)=\sec ^{2} t+\sec t \tan t
$$
The general anti-derivative of $
f^{\prime}(t)=\sec t(\sec t+\tan t)
$ is
$$
f(t)=\tan t+\sec t+C
$$
To determine C we use the fact that $f(\frac{\pi}{4})=-1$:
$$
f(\frac{\pi}{4})=\tan (\frac{\pi}{4})+\sec(\frac{\pi}{4})+C =-1
$$
$ \Rightarrow $
$$
1+\sqrt {2}+C=-1\quad \Rightarrow \quad C=-2-\sqrt {2},
$$
so
$$
f(t)=\tan t+\sec t-2-\sqrt {2}
$$
