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: $180^{\circ}-\theta $
c) Quadrant -III: $\theta - 180^o $
d) Quadrant -IV: $360^{\circ}-\theta $
Now, Step (a) refers to: $=\dfrac{\pi}{6}$
Reference angle is equal to
Since, $\tan (\dfrac{\pi}{6})=\dfrac{\sqrt 3}{3}$
So, $\tan \dfrac{-\pi}{6}=-\dfrac{\sqrt 3}{3}$; Because $\theta $ lies in Quadrant-IV.
