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: 53

Answer

$2~cos~A-sec~A = cos~A - \frac{tan~A}{csc~A}$

Work Step by Step

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