Question

Prove that :
$\frac{1+\sec A}{\sec A}=\frac{{\sin }^{2}A}{1-\cos A}$ angle A is an acute angle.

Answer 1

To prove:- $\frac{1+\sec A}{\sec A}=\frac{{\sin }^{2}A}{1-\cos A}$

Proof:-

$\frac{1+\sec A}{\sec A}$

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

$=1+\cos A$

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

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

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

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

Hence proved.