Answer
See the explanation below.
Work Step by Step
$\tan t+\frac{\cos t}{1+\sin t}=\sec t$
Recall Trigonometric Identities,
$\begin{align}
& \tan t=\frac{\sin t}{\cos t} \\
& {{\sin }^{2}}t+{{\cos }^{2}}t=1 \\
\end{align}$
Use the above identities and solve the left side of the given expression,
$\tan t+\frac{\cos t}{1+\sin t}=\frac{\sin t}{\cos t}+\frac{\cos t}{1+\sin t}$
Multiply numerator and denominator of $\tan t$ by $1+\sin t$ and $\frac{\cos t}{1+\sin t}$ by $\cos t$.
$\begin{align}
& \tan t+\frac{\cos t}{1+\sin t}=\frac{\sin t}{\cos t}\left( \frac{1+\sin t}{1+\sin t} \right)+\frac{\cos t}{1+\sin t}\left( \frac{\cos t}{\cos t} \right) \\
& =\frac{\sin t+{{\sin }^{2}}t}{\cos t+\cos t\sin t}+\frac{{{\cos }^{2}}t}{\cos t+\sin t\cos t} \\
& =\frac{\sin t+1}{\cos t\left( 1+\sin t \right)} \\
& =\frac{1}{\cos t}
\end{align}$
Recall Reciprocity Identity,
$\sec t=\frac{1}{\cos t}$
Therefore,
$\begin{align}
& \tan t+\frac{\cos t}{1+\sin t}=\frac{1}{\cos t} \\
& =\sec t
\end{align}$
