Question

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

Answer 1

$$\begin{gathered} \tan x+\cot \left(\mathrm{\pi }+x\right)+\cot \left(\frac{\mathrm{\pi }}{2}+x\right)+\cot \left(2\mathrm{\pi }-x\right)=\tan x+\left(+\cot x\right)+\cot \left(\frac{\mathrm{\pi }}{2}+x\right)+\cot \left(2\mathrm{\pi }-x\right) \\ \mathrm{Since}\cot \left(\mathrm{\pi }+x\right)=\frac{1}{\tan \left(\mathrm{\pi }+x\right)}=\frac{1}{+\tan x}=+\cot x \\ \mathrm{and}\cot \left(\frac{\mathrm{\pi }}{2}+x\right)=\frac{1}{\tan \left(\frac{\mathrm{\pi }}{2}+x\right)}=\frac{1}{-cotx}=-\tan x \\ \mathrm{and}\cot \left(2\pi -x\right)=\frac{1}{\tan \left(2\mathrm{\pi }-x\right)}=\frac{1}{-\tan x}=-\cot x \\ =\tan x+\cot x-\tan x-\cot x=0 \\ \mathrm{Hence},\tan x+\cot \left(\pi +x\right)+\cot \left(\frac{\mathrm{\pi }}{2}+x\right)+\cot \left(2\mathrm{\pi }-x\right)=0 \end{gathered}$$