#### Answer

$\dfrac{\sqrt 3}{3}$

#### Work Step by Step

The reference angle of an angle $0 \leq \theta \lt 2\pi $ based on its position can be computed by using the following steps:
a) Quadrant- I: $\theta $
b) Quadrant -II: $(\pi-\theta)$
c) Quadrant- III: $(\theta - \pi)$
d) Quadrant - IV: $(2\pi - \theta)$
We can see that the angle $\dfrac{17\pi}{6}$ is close to $\dfrac{18\pi}{6}=3\pi $.
Thus, the reference angle of $\dfrac{17\pi}{3}$ is:
$ 3 \pi- \dfrac{17\pi}{6}=\dfrac{\pi}{6}$
$\implies \tan \dfrac{\pi}{6} =\dfrac{\sqrt 3}{3}$
So, $\tan (\dfrac{-17\pi}{6})=\dfrac{\sqrt 3}{3}$ because $\theta $ lies in Quadrant-III.