Answer
$\dfrac {\cot ^{2}t-1}{1+\cot ^{2}t}=1-\sin ^{2}t-\sin ^{2}t=1-2\sin ^{2}t$
Work Step by Step
$\dfrac {\cot ^{2}t-1}{1+\cot ^{2}t}=\dfrac {\dfrac {\cos ^{2}t}{\sin ^{2}t}-1}{1+\dfrac {\cos ^{2}t}{\sin ^{2}t}}=\dfrac {\cos ^{2}t-\sin ^{2}t}{\cos ^{2}t+\sin ^{2}t}=\cos ^{2}t-\sin ^{2}t=1-\sin ^{2}t-\sin ^{2}t=1-2\sin ^{2}t$
