Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 5 - Section 5.1 - Proving Identities - 5.1 Problem Set - Page 280: 93


The equation is an identity. Please see proof in "step-by-step"

Work Step by Step

With $\displaystyle \csc\theta=\frac{1}{\sin\theta},\quad$and $\sin(-\theta)=-\sin\theta$ $ LHS=\displaystyle \frac{1}{\sin\theta}-\sin\theta,\quad$ ... common denominator is $\sin \theta$ $LHS=\displaystyle \frac{1-\sin^{2}\theta}{\sin\theta}$ in the numerator, apply the Pythagorean identity $\cos^{2}\theta+\sin^{2}\theta=1$ $LHS=\displaystyle \frac{\cos^{2}\theta+\sin^{2}\theta-\sin^{2}\theta}{\sin\theta}=\frac{\cos^{2}\theta}{\sin\theta}=RHS$
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