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