Answer
$\dfrac{1+\sin{\theta}}{\cos{\theta}}$.
Work Step by Step
Multiply to obtain:
\begin{align*}
\frac{\cos{\theta}}{1-\sin{\theta}}\cdot\frac{1+\sin{\theta}}{1+\sin{\theta}}&=\frac{\cos{\theta}(1+\sin{\theta})}{(1-\sin{\theta})(1+\sin{\theta})}\\\\&=\frac{\cos{\theta}(1+\sin{\theta})}{1-\sin^2{\theta}}
\end{align*}
We know that $\cos^2{\theta}+\sin^2{\theta}=1\longrightarrow \cos^2{\theta}=1-\sin^2{\theta}$.
Thus,
\begin{align*}
\dfrac{\cos{\theta}(1+\sin{\theta})}{1-\sin^2{\theta}}&=\frac{\cos{\theta}(1+\sin{\theta})}{\cos^2{\theta}}\\\\
&=\frac{1+\sin{\theta}}{\cos{\theta}}
\end{align*}.
