Prove the following trigonometric identities.

We need to prove $\frac{1+\cos A}{\sin ^{2} A}=\frac{1}{1-\cos A}$

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

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

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

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

$=\frac{1}{1-\cos A}$

   = RHS

Hence proved.