Answer
$-\frac{24}{7}$
Work Step by Step
Since $\theta$ is in Quadrant I and $cos\theta=\frac{3}{5}$, we have
$tan\theta=\frac{4}{3}$ and $tan2\theta=\frac{2tan\theta}{1-tan^2\theta}=\frac{8/3}{1-16/9}=-\frac{24}{7}$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 7 - Section 7.3 - Double-Angle, Half-Angle, and Product-Sum Formulas - 7.3 Exercises - Page 561: 54
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