Question

For an acute angle A , prove that
$1+{\tan }^{2}A={\sec }^{2}A$

Answer 1

$LHS=1+{\tan }^{2}A=1+\frac{{\sin }^{2}A}{{\cos }^{2}A}=\frac{{\sin }^{2}A+{\cos }^{2}A}{{\cos }^{2}A}=\frac{1}{{\cos }^{2}A}={\text{sec}}^{2}A=RHS$