Prove the following trigonometric identities.

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

Now, multiplying the numerator and denominator of LHS by $1-\cos A$, we get

$\frac{1+\cos A}{\sin A}=\frac{1+\cos A}{\sin A} \times \frac{1-\cos A}{1-\cos A}$

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

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

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

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

Hence proved.