Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.4 Sum and Difference Identities for Sine and Tangent - 5.4 Exercises - Page 228: 70


$$\frac{\tan(\alpha+\beta)-\tan\beta}{1+\tan(\alpha+\beta)\tan\beta}=\tan\alpha$$ The equation is verified to be an identity.

Work Step by Step

$$\frac{\tan(\alpha+\beta)-\tan\beta}{1+\tan(\alpha+\beta)\tan\beta}=\tan\alpha$$ If we look closely, we might realize that the left side is actually the result of an identity involving the difference of tangent of $(\alpha+\beta)$ and $\beta$. In other words,$$\frac{\tan(\alpha+\beta)-\tan\beta}{1+\tan(\alpha+\beta)\tan\beta}=\tan[(\alpha+\beta)-\beta]=\tan\alpha$$ The equation is therefore verified to be an identity.
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