Question

The minimum value of $\frac{1}{3}(3\tan x+27\cot x)$ is.......

Answer 1

$\frac{1}{3}(3\tan x+27\cot x)=\tan x+9\cot x$

We know, $AM\ge GM\Rightarrow \frac{\tan x+9\cot x}{2}\ge \sqrt{\tan x\ast 9\cot x}\Rightarrow \frac{\tan x+9\cot x}{2}\ge \sqrt{9}\Rightarrow \frac{\tan x+9\cot x}{2}\ge 3\Rightarrow \tan x+9\cot x\ge 6$

hence, its minimum value will be $6$