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 - Summary Exercises on Verifying Trigonometric Identities - Page 245: 8


$$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ The equation has been verified to be an identity as below.

Work Step by Step

$$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ We examine from the left side. $$X=\frac{2}{1+\cos x}-\tan^2\frac{x}{2}$$ $$X=\frac{2}{1+\cos x}-\Big(\tan\frac{x}{2}\Big)^2$$ From the half-angle identity for tangent: $$\tan\frac{x}{2}=\frac{\sin x}{1+\cos x}$$ Therefore, $$\Big(\tan\frac{x}{2}\Big)^2=\frac{\sin^2x}{(1+\cos x)^2}$$ Apply back to $X$: $$X=\frac{2}{1+\cos x}-\frac{\sin^2x}{(1+\cos x)^2}$$ $$X=\frac{2(1+\cos x)-\sin^2 x}{(1+\cos x)^2}$$ $$X=\frac{2+2\cos x-\sin^2x}{(1+\cos x)^2}$$ We can rewrite $\sin^2x$ into $1-\cos^2x$ according to Pythagorean identities. $$X=\frac{2+2\cos x-(1-\cos^2x)}{(1+\cos x)^2}$$ $$X=\frac{2+2\cos x-1+\cos^2x}{(1+\cos x)^2}$$ $$X=\frac{1+2\cos x+\cos^2x}{(1+\cos x)^2}$$ $$X=\frac{(1+\cos x)^2}{(1+\cos x)^2}$$ $$X=1$$ That concludes $$\frac{2}{1+\cos x}-\tan^2\frac{x}{2}=1$$ 2 sides are equal, so the equation is an identity.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.