Question

If $\tan \left(\cot x\right)=\cot \left(\tan x\right)$, then the least positive value of $\cot x+\tan x$ is

• A. $\frac{\pi }{2}$
• B. $\frac{3\pi }{2}$
• C. $2$
• D. $1$

Answer 1

The correct option is B $\frac{3\pi }{2}$

$\tan \left(\cot x\right)=\cot \left(\tan x\right)$
$\Rightarrow \tan \left(\cot x\right)=\tan (\frac{\pi }{2}-\tan x)$
$\Rightarrow \cot x=n\pi +\frac{\pi }{2}-\tan x,n\in Z$
$\Rightarrow \cot x+\tan x=n\pi +\frac{\pi }{2}$
$\Rightarrow \cot x+\tan x=(2n+1)\frac{\pi }{2}$
So, positive values of $\tan x+\cot x$ are $\{\frac{\pi }{2},\frac{3\pi }{2},...\}$

We know that for positive numbers $a,b$
$a+b\ge 2\sqrt{ab}$
$\tan x+\cot x\ge 2\sqrt{\tan x\cdot \cot x}\Rightarrow \tan x+\cot x\ge 2$

As $\frac{\pi }{2}<2$, so the least positive value of $\cot x+\tan x$ is $\frac{3\pi }{2}$