# The value of tan x+cot

Question:

The value of $\tan x+\cot (\pi+x)+\cot \left(\frac{\pi}{2}+x\right)+\cot (2 \pi-x)$ is____________ .

Solution:

$\tan x+\cot (\pi+x)+\cot \left(\frac{\pi}{2}+x\right)+\cot (2 \pi-x)=\tan x+(+\cot x)+\cot \left(\frac{\pi}{2}+x\right)+\cot (2 \pi-x)$

Since $\cot (\pi+x)=\frac{1}{\tan (\pi+x)}=\frac{1}{+\tan x}=+\cot x$

and $\cot \left(\frac{\pi}{2}+x\right)=\frac{1}{\tan \left(\frac{\pi}{2}+x\right)}=\frac{1}{-\cot x}=-\tan x$

and $\cot (2 \pi-x)=\frac{1}{\tan (2 \pi-x)}=\frac{1}{-\tan x}=-\cot x$

$=\tan x+\cot x-\tan x-\cot x=0$

Hence, $\tan x+\cot (\pi+x)+\cot \left(\frac{\pi}{2}+x\right)+\cot (2 \pi-x)=0$