Question

Prove that $\frac{1+\sec A}{\sec A}=\frac{{\sin }^{2}A}{1-\cos A}$.

Answer 1

To prove : $\frac{1+secA}{secA}=\frac{se{c}^{12}A}{1-cosA}$

LHS $\frac{1+secA}{secA}=\frac{1+\frac{1}{cosA}}{\frac{1}{cosA}}$

$=\frac{\frac{cosA+1}{cosA}}{\frac{1}{cosA}}$

$=\frac{cosA+1}{cosA}\times \frac{cosA}{1}$

$=1+cosA$

$=1+cosA\times \frac{1-cosA}{1-cosA}$

$=\frac{(1{)}^2-(cosA{)}^2}{1-cosA}$

$=\frac{1-co{s}^{2}A}{1-cosA}$

$=\frac{si{n}^{2}A}{1-cosA}$

$=RHS$