149
Prove that $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1-\cos A}$.
Show SolutionHide Solution↓
$LHS = \frac{1 + \sec A}{\sec A} = \frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$
$= 1 + \cos A$
$= \frac{(1 - \cos A)(1 + \cos A)}{(1-\cos A)}$
$= \frac{1- \cos^2 A}{1-\cos A}$
$= \frac{\sin^2 A}{1-\cos A} = RHS$
$= 1 + \cos A$
$= \frac{(1 - \cos A)(1 + \cos A)}{(1-\cos A)}$
$= \frac{1- \cos^2 A}{1-\cos A}$
$= \frac{\sin^2 A}{1-\cos A} = RHS$