Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 39

Answer

$\tan{\theta}=\dfrac{\sqrt{3}}{3}$ $\csc{\theta} =2$ $\sec{\theta} =\dfrac{2\sqrt{3}}{3}$ $\cot{\theta} =\sqrt{3}$

Work Step by Step

$\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$ $\therefore \tan{\theta}= \dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=\dfrac{\sqrt{3}}{3}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\therefore \csc{\theta} = \dfrac{1}{\dfrac{1}{2}}=2$ $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ $\therefore \sec{\theta} = \dfrac{1}{\dfrac{\sqrt{3}}{2}}= \dfrac{2\sqrt{3}}{3}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{\dfrac{\sqrt{3}}{3}}= \sqrt{3}$
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