Question

Prove: $({\sec }^{2}A+{\tan }^{2}A)(co{\sec }^{2}A+{\cot }^{2}A)=1+2{\sec }^{2}Aco{\sec }^{2}A$

Answer 1

$LHS=({\text{sec}}^{2}A+{\tan }^{2}A)({\text{cosec}}^{2}A+{\cot }^{2}A)$

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

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

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

$LHS=RHS$

Hence Proved