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: 94

Answer

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

Work Step by Step

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