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: 54



Work Step by Step

$cos^2v+sin^2v=1$ $\frac{16}{25}+sin^2v=1$ $sin^2v=\frac{9}{25}$ $sin~v=±\frac{3}{5}~~$ (But, $v$ is in Quadrant III): $sin~v=-\frac{3}{5}$ $tan~v=\frac{sin~v}{cos~v}=\frac{-\frac{3}{5}}{-\frac{4}{5}}=\frac{3}{5}·\frac{5}{4}=\frac{3}{4}$ $tan(u+v)=\frac{tan~u+tan~v}{1-tan~u~tan~v}=\frac{\frac{3}{4}+\frac{3}{4}}{1-\frac{3}{4}·\frac{3}{4}}=\frac{\frac{3}{2}}{1-\frac{9}{16}}=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{3}{2}·\frac{16}{7}=\frac{24}{7}$
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