Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 5 - Review - Exercises - Page 463: 26

Answer

$0.9916$

Work Step by Step

Given: $\sin t=\dfrac{-8}{17}$ This implies that $\csc t=\dfrac{1}{\sin t}=-\dfrac{17}{8}$ and $\cos t=\sqrt {1-\sin^2 t} \implies \cos t=\sqrt {1-(\dfrac{-8}{17})^2 }$ or, $\cos t=\dfrac{15}{17}$ As $t$ lies in the fourth quadrant, thus the value of $\cos$ will be positive. That is, $\sec t=\dfrac{1}{\cos t}=\dfrac{17}{15}$ Now, $\csc t+\sec t=-\dfrac{17}{8}+\dfrac{\sqrt {17}}{15}=0.9916$
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