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: 55


$1+tan^2~\alpha = 2~tan~\alpha ~csc~2\alpha$

Work Step by Step

$1+tan^2~\alpha = 1+\frac{sin^2~\alpha}{cos^2~\alpha}$ $1+tan^2~\alpha = \frac{cos^2~\alpha}{cos^2~\alpha}+\frac{sin^2~\alpha}{cos^2~\alpha}$ $1+tan^2~\alpha = \frac{cos^2~\alpha + sin^2~\alpha}{cos^2~\alpha}$ $1+tan^2~\alpha = \frac{1}{cos^2~\alpha}$ $1+tan^2~\alpha = \frac{2~sin~\alpha}{2~sin~\alpha~cos^2~\alpha}$ $1+tan^2~\alpha = \frac{2~sin~\alpha}{2~sin~\alpha~cos~\alpha~cos~\alpha}$ $1+tan^2~\alpha = \frac{2~sin~\alpha}{sin~2\alpha~cos~\alpha}$ $1+tan^2~\alpha = 2~tan~\alpha ~csc~2\alpha$
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