Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - 7.1 Fundamental Identities - 7.1 Exercises: 31

Answer

$csc\theta =\dfrac {3}{2};$ $\cos \theta =-\dfrac {\sqrt {5}}{3};$ $sec\theta =-\dfrac {3}{\sqrt {5}};$ $\tan \theta =-\dfrac {2}{\sqrt {5}};$ $\cot \theta =-\dfrac {\sqrt {5}}{2}$

Work Step by Step

In quadrant 2 only $\csc \theta$ and $\sin\theta$ are positive other 4 trigonometric functions are negative . Then $csc\theta =\dfrac {1}{\sin \theta }=\dfrac {\dfrac {1}{2}}{3}=\dfrac {3}{2};$ $\cos \theta =-\sqrt {1-\sin ^{2}\theta =}-\sqrt {1-\left( \dfrac {2}{3}\right) ^{2}}=-\dfrac {\sqrt {5}}{3};$ $sec\theta =\dfrac {1}{\cos \theta }=\dfrac {1}{-\dfrac {\sqrt {5}}{3}}=-\dfrac {3}{\sqrt {5}};$ $\tan \theta =\dfrac {\sin \theta }{\cos \theta }=\dfrac {\dfrac {2}{3}}{-\dfrac {\sqrt {5}}{3}}=-\dfrac {2}{\sqrt {5}};$ $\cot \theta =\dfrac {1}{\tan \theta }=-\dfrac {\sqrt {5}}{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.