Answer
$-\dfrac{\sqrt3}{3}$
Work Step by Step
RECALL:
In a unit circle
$\cos {t}=x
\\\sin{t}=y
\\\tan{t} = \dfrac{y}{x}, x\ne0
\\\cot{t} = \dfrac{x}{y}, y \ne 0
\\\sec{t} = \dfrac{1}{x}, x\ne0
\\\csc{t} = \dfrac{1}{y}, y \ne0$
Using the definition above and the unit circle in Figure 5 on page 137 of this book, then:
Point on the unit circle: $\left(-\dfrac{\sqrt3}{2}, \dfrac{1}{2}\right)$
Thus,
$\tan{(\frac{5\pi}{6})} = \dfrac{y}{x}=\dfrac{\frac{1}{2}}{-\frac{\sqrt3}{2}}=\dfrac{1}{2} \cdot \dfrac{-2}{\sqrt{3}}=-\dfrac{1}{\sqrt3}$
Rationalize the denominator to obtain:
$\tan{(\frac{5\pi}{6})}=-\dfrac{1}{\sqrt3} \cdot \dfrac{\sqrt3}{\sqrt3}=-\dfrac{\sqrt3}{3}$
