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

Published by Pearson

# Chapter 5 - Trigonometric Functions - Section 5.3 Properties of the Trigonometric Functions - 5.3 Assess Your Understanding - Page 418: 38

#### Answer

$\tan{\theta}=\dfrac{1}{2}$ $\csc{\theta} =-\sqrt{5}$ $\sec{\theta} =-\dfrac{\sqrt{5}}{2}$ $\cot{\theta} =2$

#### Work Step by Step

$\tan{\theta}= \dfrac{\sin{\theta}}{\cos{\theta}}$ $\therefore \tan{\theta}= \dfrac{-\dfrac{\sqrt{5}}{5}}{-\dfrac{2\sqrt{5}}{5}}=\dfrac{1}{2}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}}$ $\therefore \csc{\theta} = \dfrac{1}{-\dfrac{\sqrt{5}}{5}}=-\sqrt{5}$ $\sec{\theta} = \dfrac{1}{\cos{\theta}}$ $\therefore \sec{\theta} = \dfrac{1}{-\dfrac{2\sqrt{5}}{5}}= -\dfrac{\sqrt{5}}{2}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}$ $\cot{\theta} = \dfrac{1}{\dfrac{1}{2}}= 2$

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