Answer
$\color{blue}{\tan{(\frac{5\pi}{6})}=-\dfrac{\sqrt3}{3}}$
Work Step by Step
Figure 13 on page 579 of this book shows that the unit circle point $(-\frac{\sqrt3}{2}, \frac{1}{2})$ corresponds to the real number $\dfrac{5\pi}{6}$.
RECALL:
(i) $\cos{s} = x$
(ii) $\sin{s} = y$
(iii) $\tan{s}=\dfrac{y}{x}$
(iv) $\sec{s} =\dfrac{1}{x}$
(v) $\csc{s} = \dfrac{1}{y}$
(vi) $\cot{s} = \dfrac{x}{y}$
Use the coordinates of the unit circle point above and the formula in (iii) above to obtain:
$\color{blue}{\tan{(\frac{5\pi}{6})}=\dfrac{\frac{1}{2}}{-\frac{\sqrt3}{2}}=\frac{1}{2} \cdot \frac{-2}{\sqrt3}=-\frac{1}{\sqrt3}=-\frac{1\sqrt3}{(\sqrt3)(\sqrt3)}=-\frac{\sqrt3}{3}}$
