Question

If $x,y,z$ be all positive acute angles, then the least value of
$\tan x(\cot y+\cot z)+\tan y(\cot z+\cot x)+\tan z(\cot x+\cot y),$ is

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is C $6$

If a number $m$ is positive so

$m+\frac{1}{m}\ge 2$

So, we will use this rule here,

$\tan x(\cot y+\cot z)+\tan y(\cot z+\cot x)+\tan z(\cot x+\cot y)$

$=\tan x(\frac{1}{\tan y}+\frac{1}{\tan z})+\tan y(\frac{1}{\tan z}+\frac{1}{\tan x})+\tan z(\frac{1}{\tan x}+\frac{1}{\tan y})$

$=\frac{\tan x}{\tan y}+\frac{\tan x}{\tan z}+\frac{\tan y}{\tan z}+\frac{\tan y}{\tan x}+\frac{\tan z}{\tan x}+\frac{\tan z}{\tan y}$

$=(\frac{\tan x}{\tan y}+\frac{\tan y}{\tan x})+(\frac{\tan x}{\tan z}+\frac{\tan z}{\tan x})+(\frac{\tan y}{\tan z}+\frac{\tan z}{\tan y})$

$=(a+\frac{1}{a})+(b+\frac{1}{b})+(c+\frac{1}{c})$

where $a=\frac{\tan x}{\tan y},b=\frac{\tan y}{\tan z},c=\frac{\tan z}{\tan x}$

$\ge 2+2+2$(from above)

$\ge 6$

So, minimum value is $6$