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 237: 38



Work Step by Step

$$X=\frac{2\tan15^\circ}{1-\tan^215^\circ}$$ - From Double-Angle Identity for tangent: $$\tan2A=\frac{2\tan A}{1-\tan^2A}$$ So if we replace the above identity with $A=15^\circ$ as in $X$, we get $$X=\tan(2\times15^\circ)$$ $$X=\tan30^\circ$$ $$X=\frac{1}{\sqrt3}=\frac{\sqrt3}{3}$$ Therefore, $$\frac{2\tan15^\circ}{1-\tan^215^\circ}=\frac{\sqrt3}{3}$$
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