Question

Prove that $\frac{secA-tanA}{secA+tanA}=\frac{co{s}^{2}A}{(1+sinA{)}^2}$

Answer 1

$LHS=\frac{secA-tanA}{secA+tanA}$

Multiply and divide LHS by $(\sec A+\tan A)$

$=\frac{{\sec }^{2}A-{\tan }^{2}A}{(\sec A+\tan A{)}^2}$

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

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

$=RHS$

Hence proved