Answer
$\dfrac {1-\sin ^{2}\theta }{\sin ^{2}\theta }=\dfrac {1}{\sin ^{2}\theta }-1$
Work Step by Step
$csc^{2}\theta \cos ^{2}\theta =\dfrac {1}{\sin ^{2}\theta }\cos ^{2}\theta =\dfrac {1-\sin ^{2}\theta }{\sin ^{2}\theta }=\dfrac {1}{\sin ^{2}\theta }-1$
