Answer
$\dfrac{1+\sin{\theta}}{\cos{\theta}}$
Work Step by Step
Recall:
$(a-b)(a+b)=a^2-b^2$
Use the rule above to obtain:
\begin{align*}
\dfrac{\cos{\theta}}{1-\sin{\theta}} \times \dfrac{1+\sin{\theta}}{1+\sin{\theta}}&= \dfrac{\cos{\theta}(1+\sin{\theta})}{(1-\sin{\theta})(1+\sin{\theta})}\\\\
&= \dfrac{\cos{\theta}(1+\sin{\theta})}{1-\sin^2{\theta}}
\end{align*}
Since $1-\sin^2{\theta} = \cos^2{\theta}$, then:
$\dfrac{\cos{\theta}(1+\sin{\theta})}{1-\sin^2{\theta}} = \dfrac{\cos{\theta}(1+\sin{\theta})}{\cos^2{\theta}} = \boxed{\dfrac{1+\sin{\theta}}{\cos{\theta}}}$
