Answer
$\sin ^{2}\theta \left( 1+6t^{2}\theta \right) -1=\\sin ^{2}\theta +\cos ^{2}\theta -1=1-1=0$
Work Step by Step
$\sin ^{2}\theta \left( 1+6t^{2}\theta \right) -1=\sin ^{2}\theta +\sin ^{2}\theta \cot ^{2}\theta -1=\sin ^{2}\theta +\sin ^{2}\theta \dfrac {\cos ^{2}\theta }{\sin ^{2}\theta }-1=\sin ^{2}\theta +\cos ^{2}\theta -1=1-1=0$
