Question

Let $\left(x\right)=\frac{1+\cos 2x+8{\sin }^{2}x}{\sin 2x},x\in \left(0,\frac{\pi }{2}\right).$ find minimum value of $f\left(x\right)$

Answer 1

$f\left(x\right)=\frac{1+\cos 2x+8{\sin }^{2}x}{\sin 2x}=\frac{{\cos }^{2}x+4{\sin }^{2}x}{\sin x\cos x}=\cot x+4\tan x$

$\frac{\cot x+4\tan x}{2}\ge \sqrt{4}⟹f\left(x\right)\ge 4$

Minimum value of $f\left(x\right)$ is $4$