Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Test - Page 244: 5b

Answer

$tan(\pi+x) = tan~x$

Work Step by Step

$tan(\pi+x) = \frac{tan~\pi+tan~x}{1-tan~\pi~tan~x}$ $tan(\pi+x) = \frac{\frac{sin~\pi}{cos~\pi}+\frac{sin~x}{cos~x}}{1-(\frac{sin~\pi}{cos~\pi})~(\frac{sin~x}{cos~x})}$ $tan(\pi+x) = \frac{\frac{0}{cos~\pi}+\frac{sin~x}{cos~x}}{1-(\frac{0}{cos~\pi})~(\frac{sin~x}{cos~x})}$ $tan(\pi+x) = \frac{\frac{sin~x}{cos~x}}{1}$ $tan(\pi+x) = tan~x$
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