Answer
$\dfrac {\sin t}{1+\cos t}=\dfrac {1-\cos t}{\sin t}$
Work Step by Step
$\dfrac {\sin t}{1+\cos t}=\dfrac {\sin t\left( 1-\cos t\right) }{\left( 1+\cos t\right) \left( 1-\cos t\right) }=\dfrac {\sin t\left( 1-\cos t\right) }{\sin ^{2}t}=\dfrac {1-\cos t}{\sin t}$
