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