Prove the following trigonometric identities.

In the given question, we need to prove $\frac{1-\tan ^{2} A}{\cot ^{2} A-1}=\tan ^{2} A$

Now, using $\tan \theta=\frac{\sin \theta}{\cos \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$ in the L.H.S, we get

$\frac{1-\tan ^{2} A}{\cot ^{2} A-1}=\frac{1-\frac{\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A}{\sin ^{2} A}-1}$

$=\frac{\frac{\cos ^{2} A-\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A-\sin ^{2} A}{\sin ^{2} A}}$

$=\frac{\cos ^{2} A-\sin ^{2} A}{\cos ^{2} A} \times \frac{\sin ^{2} A}{\cos ^{2} A-\sin ^{2} A}$

Solving further, we get

$\frac{\cos ^{2} A-\sin ^{2} A}{\cos ^{2} A} \times \frac{\sin ^{2} A}{\cos ^{2} A-\sin ^{2} A}=\frac{\sin ^{2} A}{\cos ^{2} A}$

$=\tan ^{2} A$

Hence proved.