Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Review Exercises - Page 243: 63

Answer

$\frac{sin^2~x-cos^2~x}{csc~x} = 2~sin^3~x-sin~x$

Work Step by Step

$\frac{sin^2~x-cos^2~x}{csc~x} = \frac{sin^2~x-cos^2~x}{\frac{1}{sin~x}}$ $\frac{sin^2~x-cos^2~x}{csc~x} = sin^3~x-sin~x~cos^2~x$ $\frac{sin^2~x-cos^2~x}{csc~x} = sin^3~x-sin~x~(1-sin^2~x)$ $\frac{sin^2~x-cos^2~x}{csc~x} = sin^3~x-sin~x~+sin^3~x$ $\frac{sin^2~x-cos^2~x}{csc~x} = 2~sin^3~x-sin~x$
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