Answer
$-\frac{\sqrt3}{3}$
Work Step by Step
RECALL:
$\sin{s} = y
\\\cos{s} = x
\\\tan{s} = \frac{y}{x}
\\\cot{s} = \frac{x}{y}
\\\sec{s} = \frac{1}{x}
\\\csc{s}=\frac{1}{y}$
(refer to Figure 11 , page 111 of the textbook)
This angle intersects the unit circle at the point $(-\frac{\sqrt3}{2}, \frac{1}{2})$.
This point has:
$x= -\frac{\sqrt3}{2}$
$y=\frac{1}{2}$
Thus,
$\tan{\frac{5\pi}{6}}
\\= \dfrac{y}{x}
\\=\dfrac{\frac{1}{2}}{-\frac{\sqrt3}{2}}
\\=\frac{1}{2} \cdot (-\frac{2}{\sqrt3})
\\=-\frac{1}{\sqrt3}$
Rationalize the denominator by multiplying $\sqrt3$ to both the numerator and the denominator to obtain:
$=-1 \cdot \frac{\sqrt3}{\sqrt3}
\\=-\frac{\sqrt3}{3}$