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: 54


$\frac{2~tan~B}{sin~2B} = sec^2~B$

Work Step by Step

$\frac{2~tan~B}{sin~2B} = \frac{2~\frac{sin~B}{cos~B}}{2~sin~B~cos~B}$ $\frac{2~tan~B}{sin~2B} = \frac{\frac{1}{cos~B}}{cos~B}$ $\frac{2~tan~B}{sin~2B} = \frac{1}{cos^2~B}$ $\frac{2~tan~B}{sin~2B} = sec^2~B$
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