#### Answer

$\sin ^{2}\theta \cos ^{2}\theta $

#### Work Step by Step

$\dfrac {1-\cos ^{2}\left( -\theta \right) }{1+\tan ^{2}\left( -\theta \right) }=\dfrac {1-\left( \cos \left( -\theta \right) \right) ^{2}}{1+\left( \dfrac {\sin \left( -\theta \right) }{\cos \left( -\theta \right) }\right) ^{2}}=\dfrac {1-\left( \cos \theta \right) ^{2}}{1+\left( \dfrac {-\sin \theta }{\cos \theta }\right) ^{2}}=\dfrac {\sin ^{2}\theta }{1+\dfrac {\sin ^{2}\theta }{\cos ^{2}\theta }}=\dfrac {\sin ^{2}\theta \cos ^{2}\theta }{\cos ^{2}\theta +\sin ^{2}\theta }=\sin ^{2}\theta \cos ^{2}\theta $