Prove the following trigonometric identities.

We need to prove $\left(1+\tan ^{2} A\right)+\left(1+\frac{1}{\tan ^{2} A}\right)=\frac{1}{\sin ^{2} A-\sin ^{4} A}$.

Using the property $1+\tan ^{2} \theta=\sec ^{2} \theta$, we get

$\left(1+\tan ^{2} A\right)+\left(1+\frac{1}{\tan ^{2} A}\right)=\sec ^{2} A+\left(\frac{\tan ^{2} A+1}{\tan ^{2} A}\right)$

$=\sec ^{2} A+\left(\frac{\sec ^{2} A}{\tan ^{2} A}\right)$

Now, using $\sec \theta=\frac{1}{\cos \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$, we get

$\sec ^{2} A+\left(\frac{\sec ^{2} A}{\tan ^{2} A}\right)=\frac{1}{\cos ^{2} A}+\left(\frac{\frac{1}{\cos ^{2} A}}{\frac{\sin ^{2} A}{\cos ^{2} A}}\right)$

$=\frac{1}{\cos ^{2} A}+\left(\frac{1}{\cos ^{2} A} \times \frac{\cos ^{2} A}{\sin ^{2} A}\right)$

Now, using $\sec \theta=\frac{1}{\cos \theta}$ and $\tan \theta=\frac{\sin \theta}{\cos \theta}$, we get

$\sec ^{2} A+\left(\frac{\sec ^{2} A}{\tan ^{2} A}\right)=\frac{1}{\cos ^{2} A}+\left(\frac{\frac{1}{\cos ^{2} A}}{\frac{\sin ^{2} A}{\cos ^{2} A}}\right)$

$=\frac{1}{\cos ^{2} A}+\left(\frac{1}{\cos ^{2} A} \times \frac{\cos ^{2} A}{\sin ^{2} A}\right)$

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

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

Further, using the property, $\sin ^{2} \theta+\cos ^{2} \theta=1$, we get

$\frac{\sin ^{2} A+\cos ^{2} A}{\cos ^{2} A\left(\sin ^{2} A\right)}=\frac{1}{\cos ^{2} A\left(\sin ^{2} A\right)}$

$=\frac{1}{\left(1-\sin ^{2} A\right)\left(\sin ^{2} A\right)}$ (using $\cos ^{2} \theta=1-\sin ^{2} \theta$ )

$=\frac{1}{\left.\sin ^{2} A-\sin ^{4} A\right)}$

Hence proved.