Answer
$\cot{(\frac{2\pi}{3})}=-\frac{\sqrt3}{3}$
Work Step by Step
Convert $\dfrac{2\pi}{3}$ to degrees to obtain:
$\require{cancel}
\\=\dfrac{2\pi}{3} \times \dfrac{180^o}{\pi}
\\=\dfrac{2\cancel{\pi}}{\cancel{3}} \times \dfrac{\cancel{180^o}^{60^o}}{\cancel{\pi}}
\\=120^o$
The reference angle of $120^o$ is $60^o$.
Note that:
$\sin{60^o}= \frac{\sqrt{3}}{2}
\\\cos{60^o}=\frac{1}{2}$
However, since $\dfrac{2\pi}{3} = 120^o$ and this angle terminates in Quadrant II, then
$\sin{120^o}= \frac{\sqrt{3}}{2}
\\\cos{120^o}=-\frac{1}{2}$
RECALL:
$\cot {\theta} = \dfrac{\cos{\theta}}{\sin{\theta}}$
Therefore,
$\cot{(\frac{2\pi}{3})}
\\=\cot{120^o}
\\=\dfrac{\cos{120^o}}{\sin{120^o}}
\\=\dfrac{-\frac{1}{2}}{\frac{\sqrt3}{2}}
\\=-\frac{1}{2} \times \frac{2}{\sqrt3}
\\=-\frac{1}{\sqrt3}
\\=-\frac{1}{\sqrt3} \times \frac{\sqrt3}{\sqrt3}
\\=-\frac{\sqrt3}{3}$
