# Chapter 13, Trigonometric Ratios and Functions - 13.3 Evaluate Trigonometric Functions of Any Angle - 13.3 Exercises - Skill Practice - Page 871: 27

$-\dfrac{2 \sqrt 3}{3}$

#### Work Step by Step

When $P$ is the period of $f(x)$ then we have $f(x+p)=f(x)$ Here, we have $\csc (-420^{\circ})=-\csc 420^{\circ}=-\csc 60^{\circ} =-\dfrac{1}{\sin 60^{\circ}}$ and $-\dfrac{1}{\sin 60^{\circ}}=-\dfrac{1}{\sqrt 3/2}$ or, $=-\dfrac{2}{\sqrt 3}$ or, $=-\dfrac{2 \sqrt 3}{3}$

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