Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 7 - Trigonometric Identities and Equations - Test - Page 740: 11


$\tan^2 x-\sin^2 x=(\tan x\sin x)^2$

Work Step by Step

Start with the left side: $\tan^2 x-\sin^2 x$ Rewrite in terms of sine and cosine: $=\frac{\sin^2x}{\cos^2 x}-\sin^2 x$ Get a common denominator of $\cos^2 x$: $=\frac{\sin^2x}{\cos^2 x}-\frac{\sin^2x\cos^2x}{\cos^2 x}$ $=\frac{\sin^2x-\sin^2x\cos^2x}{\cos^2 x}$ Simplify: $=\frac{\sin^2x(1-\cos^2x)}{\cos^2 x}$ $=\frac{\sin^2x\sin^2x}{\cos^2 x}$ $=\sin^2 x*\frac{\sin^2x}{\cos^2 x}$ $=\sin^2 x\tan^2 x$ $=(\tan x\sin x)^2$ Since this equals the right side, the identity is proven.
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