Question

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

Answer 1

Consider $LHS=\frac{1+\sec A}{\sec A}$

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

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

$=(1+\cos A)$

Multiply and divide with $(1-\cos A)$

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

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

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

$=RHS$