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