Question

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

Answer 1

To prove ${\sec }^{2}x+{\csc }^{2}x\ge 4$

Let $F={\sec }^{2}x+{\csc }^{2}x$

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

$\Rightarrow F=\frac{1}{{\sin }^{2}x{\cos }^{2}x}=\frac{1}{(2\sin x\cos x{)}^2}\frac{1}{2^2}$ .................... $[sin²\theta +cos²\theta =1]$

$\Rightarrow F=\frac{4}{(\sin 2x{)}^2}$ ........ $[\because \sin 2x=2\sin x\cos x]$

We know that for any values of $\theta$

$0\le {\sin }^2\left(\theta \right)\le 1$

$\therefore F=\frac{4}{(\sin 2x{)}^2}\Rightarrow 4\le F<\infty$

i.e. lower bond on $F$ is $4$

$\therefore F={\sec }^{2}x+{\csc }^2\ge 4$