Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 7 - 7.1 - Using Fundamental Identities - 7.1 Exercises - Page 514: 56


$\sqrt {9x^2+25}=5~sec~θ$

Work Step by Step

$sec^2θ=1+tan^2θ$ $\sqrt {9x^2+25}=\sqrt {(3x)^2+25}=\sqrt {(5~tan~θ)^2+25}=\sqrt {25~tan^2θ+25}=\sqrt {25(tan^2θ+1)}=5\sqrt {sec^2θ}=±5~sec~θ$ But, since $0\ltθ\lt\frac{\pi}{2}$ (Quadrant I): $\sqrt {9x^2+25}=5~sec~θ$
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