Question

Prove that: ${\sec }^{2}x+{\mathrm{cosec}}^{2}x\ge 4$

Answer 1

Given,

${\sec }^{2}x+{\csc }^{2}x$

$=1+{\tan }^{2}x+1+{\cot }^{2}x$

$=2+{\tan }^{2}x+{\cot }^{2}x$

we know that, $AM>GM$

$\frac{{\tan }^{2}x+{\cot }^{2}x}{2}>\sqrt{{\tan }^{2}x{\cot }^{2}x}$

${\tan }^{2}x+{\cot }^{2}x>2$

${\tan }^{2}x+{\cot }^{2}x+2>2+2$

${\tan }^{2}x+{\cot }^{2}x+2>4$

$1+{\tan }^{2}x+1+{\cot }^{2}x>4$

${\sec }^{2}x+{\csc }^{2}x>4$

Hence proved.