Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{1+\sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$

(A) To prove: $\frac{1+\sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$
Step $1$: Simplify the Left Hand Side $(L.H.S.)$.
$L.H.S. = \frac{1+\sec A}{\sec A} = \frac{1 + \frac{1}{\cos A}}{\frac{1}{\cos A}}$
Step $2$: Simplify the fraction.
$= \frac{\frac{\cos A + 1}{\cos A}}{\frac{1}{\cos A}} = \cos A + 1$
Step $3$: Simplify the Right Hand Side $(R.H.S.)$ to match the $L.H.S.$ or multiply the $L.H.S.$ by $\frac{1-\cos A}{1-\cos A}$.
$= (1 + \cos A) \times \frac{1-\cos A}{1-\cos A}$
Step $4$: Use the identity $\sin^2 A + \cos^2 A = 1$,so $1 - \cos^2 A = \sin^2 A$.
$= \frac{1 - \cos^2 A}{1 - \cos A} = \frac{\sin^2 A}{1 - \cos A}$
Thus,$L.H.S. = R.H.S.$