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 5.5 Double-Angle Identities - 5.5 Exercises - Page 236: 2


$$\frac{2\tan15^\circ}{1-\tan^215^\circ}=\frac{\sqrt3}{3}$$ 2 is matched with E.

Work Step by Step

$$X=\frac{2\tan15^\circ}{1-\tan^215^\circ}$$ Recall the double-angle identities: $$\frac{2\tan A}{1-\tan^2A}=\tan2A$$ Here $X$ is like the form of $\frac{2\tan A}{1-\tan^2A}$ with $A=15^\circ$. Therefore, $$X==\frac{2\tan15^\circ}{1-\tan^215^\circ}=\tan(2\times15^\circ)$$ $$X=\tan30^\circ$$ $$X=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$$ So, $$\frac{2\tan15^\circ}{1-\tan^215^\circ}=\frac{\sqrt3}{3}$$ We pick E here.
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