Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Section 5.3 - Double-Angle, Power-Reducing, and Half-Angle Formulas - Exercise Set - Page 681: 73


See the explanation below.

Work Step by Step

We use the trigonometric identities: $2\sin x\cos x=\sin 2x$ and $\cos 2x=2{{\cos }^{2}}x-1$. Now, the left side can be written as: $\begin{align} & \frac{\sin 2x}{\sin x}-\frac{\cos 2x}{\cos x}=\frac{2\sin x\cos x}{\sin x}-\frac{\left( 2{{\cos }^{2}}x-1 \right)}{\cos x} \\ & =2\cos x-\frac{2{{\cos }^{2}}x}{\cos x}+\frac{1}{\cos x} \\ & =2\cos x-2\cos x+\sec x \\ & =\sec x \end{align}$ Thus, the left side is equal to $\sec x$.
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