Algebra and Trigonometry 10th Edition

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

Chapter 7 - Review Exercises - Page 552: 18


$\sqrt {x^2-16}=4~tan~θ$

Work Step by Step

$1+tan^2θ=sec^2θ$ $sec^2θ-1=tan^2θ$ $\sqrt {x^2-16}=\sqrt {(4~sec^2θ)^2-16}=\sqrt {16~sec^2θ-16}=\sqrt {16(sec^2θ-1)}=4\sqrt {tan^2~θ}=±4~tan~θ$ But, since $0\lt θ\lt\frac{\pi}{2}$: $\sqrt {x^2-16}=4~tan~θ$
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