Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 14 - Trigonometric Identities and Equations - 14-1 Trigonometric Identities - Practice and Problem-Solving Exercises - Page 908: 18

Answer

$$\tan^{2}\theta$$

Work Step by Step

Use the Pythagorean Identity: $$1+\tan^{2}\theta=\sec^{2}\theta$$ Rearrange terms: $$\sec^{2}\theta-1=\tan^{2}\theta$$ Alternatively, replace $\sec{\theta}$ with $\frac{1}{\cos\theta}$ $$\sec^{2}\theta-1= \frac{1}{cos^{2}\theta}-1$$ Common denominator: $$\sec^{2}\theta-1= \frac{1-cos^{2}\theta}{cos^{2}\theta}$$ By identity, $sin^{2}\theta+cos^{2}\theta=1$ (subtract $cos^{2}\theta$ on both sides) Therefore, replace 1-$cos^{2}\theta$ with $sin^{2}\theta$ $$\sec^{2}\theta-1=\frac{sin^{2}\theta}{cos^{2}\theta}$$ By identity: $\frac{sin\theta}{cos\theta}=tan\theta$ Hence, $$\sec^{2}\theta-1=tan^{2}\theta$$
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