Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.4 - Introduction to Identities - 1.4 Problem Set - Page 41: 54

Answer

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

Work Step by Step

$\sec{\theta} = b$ $\cos{\theta} = \dfrac{1}{\sec{\theta}} = \dfrac{1}{b}$ $\because \theta \in QI \hspace{20pt} \therefore \sin{\theta}$ is positive. $\sin{\theta} = \sqrt{1-\cos^2{\theta}} = - \sqrt{1-\left(\dfrac{1}{b}\right)^2} = \sqrt{\dfrac{b^2-1}{b^2}} = \dfrac{\sqrt{b^2-1}}{b}$ $\tan{\theta} = \dfrac{\sin{\theta}}{\cos{\theta}} = \sqrt{b^2-1}$ $\csc{\theta} = \dfrac{1}{\sin{\theta}} = \dfrac{b}{\sqrt{b^2-1}}$ $\cot{\theta} = \dfrac{1}{\tan{\theta}}= \dfrac{1}{\sqrt{b^2-1}} $
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