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 5.6 Half-Angle Identities - 5.6 Exercises - Page 243: 49


The equation $$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ is an identity, as 2 sides are proved to be equal below.

Work Step by Step

$$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ The left side would be examined first. $$X=\frac{2}{1+\cos x}-\tan^2\frac{x}{2}$$ - Half-angle identity for tangent: $$\tan\frac{x}{2}=\pm\sqrt{\frac{1-\cos x}{1+\cos x}}$$ (we choose this identity because here we would use $\tan^2\frac{x}{2}$) $$\tan^2\frac{x}{2}=\frac{1-\cos x}{1+\cos x}$$ (The $\pm$ sign can be removed since both $\tan^2\frac{x}{2}$ and $\frac{1-\cos x}{1+\cos x}$ are positive) Apply back to $X$, we have $$X=\frac{2}{1+\cos x}-\frac{1-\cos x}{1+\cos x}$$ $$X=\frac{2-(1-\cos x)}{1+\cos x}$$ $$X=\frac{2-1+\cos x}{1+\cos x}$$ $$X=\frac{1+\cos x}{1+\cos x}$$ $$X=1$$ So the left side is equal to the right side. The equation $$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ is therefore an identity.
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