Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Test - Page 250: 5b


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

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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