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