Answer
$\cot^{2}t-\cos^{2}t=\cot^{2}t\cos^{2}t$
Work Step by Step
$\cot^{2}t-\cos^{2}t=\cot^{2}t\cos^{2}t$
Substitute $\cot^{2}t$ with $\dfrac{\cos^{2}t}{\sin^{2}t}$:
$\dfrac{\cos^{2}t}{\sin^{2}t}-\cos^{2}t=\cot^{2}t\cos^{2}t$
Take out common factor $\cos^{2}t$:
$\cos^{2}t\Big(\dfrac{1}{\sin^{2}t}-1\Big)=\cot^{2}t\cos^{2}t$
Evaluate $\dfrac{1}{\sin^{2}t}-1$:
$\cos^{2}t\Big(\dfrac{1-\sin^{2}t}{\sin^{2}t}\Big)=\cot^{2}t\cos^{2}t$
Since $1-\sin^{2}t=\cos^{2}t$, the identity is proved:
$\Big(\dfrac{\cos^{2}t}{\sin^{2}t}\Big)\cos^{2}t=\cot^{2}t\cos^{2}t$
$\cot^{2}t\cos^{2}t=\cot^{2}t\cos^{2}t$
