Question

Prove that ${\sec }^{2}x+{\mathrm{cosec}}^{2}x={\sec }^{2}x\cdot {\mathrm{cosec}}^{2}x$

Answer 1

$LHS={\text{sec}}^{2}x+{\text{cosec}}^{2}x=\frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}=\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}=\frac{1}{{\sin }^{2}x{\cos }^{2}x}=\frac{1}{{\sin }^{2}x}\cdot \frac{1}{{\cos }^{2}x}={\text{sec}}^{2}x\cdot {\text{cosec}}^{2}x$