# Chapter 4 - Section 4.4 - Trigonometric Functions of Any Angle - Exercise Set - Page 575: 76

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

#### 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.

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