Trigonometry (11th Edition) Clone

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

Chapter 5 - Review Exercises - Page 249: 58

Answer

$csc~A~sin~2A - sec~A = cos~2A~sec~A$

Work Step by Step

$csc~A~sin~2A - sec~A = \frac{2~sin~A~cos~A}{sin~A} - \frac{1}{cos~A}$ $csc~A~sin~2A - sec~A = 2~cos~A - \frac{1}{cos~A}$ $csc~A~sin~2A - sec~A = \frac{2~cos^2~A}{cos~A} - \frac{cos^2~A+sin^2~A}{cos~A}$ $csc~A~sin~2A - sec~A = \frac{cos^2~A -sin^2~A}{cos~A}$ $csc~A~sin~2A - sec~A = \frac{cos~2A}{cos~A}$ $csc~A~sin~2A - sec~A = cos~2A~sec~A$
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