Prove the following trigonometric identities.

In the given question, we need to prove $\tan ^{2} A+\cot ^{2} A=\sec ^{2} A \operatorname{cosec}^{2} A-2$

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

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

$=\frac{\sin ^{4} A+\cos ^{4} A}{\cos ^{2} A \sin ^{2} A}$

$=\frac{\left(\sin ^{2} A\right)^{2}+\left(\cos ^{2} A\right)^{2}}{\cos ^{2} A \sin ^{2} A}$

Further, using the identity $a^{2}+b^{2}=(a+b)^{2}-2 a b$, we get

$\frac{\left(\sin ^{2} A\right)^{2}+\left(\cos ^{2} A\right)^{2}}{\cos ^{2} A \sin ^{2} A}=\frac{\left(\sin ^{2} A+\cos ^{2} A\right)^{2}-2 \sin ^{2} A \cos ^{2} A}{\sin ^{2} A \cos ^{2} A}$

$=\frac{(1)^{2}-2 \sin ^{2} A \cos ^{2} A}{\sin ^{2} A \cos ^{2} A}$

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

$=\operatorname{cosec}^{2} A \sec ^{2} A-2$

Since L.H.S = R.H.S

Hence proved.