Question

Prove that:$1+\frac{ta{n}^{2}A}{1+secA}=secA.$

Answer 1

$1+\frac{{\tan }^{2}A}{1+\sec A}=1+\frac{({\sec }^{2}A-1)}{1+\sec A}$

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

$=1+\sec A-1$

$=\sec A$

$\Rightarrow 1+\frac{{\tan }^{2}A}{1+\sec A}=\sec A$.