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: 74

Answer

See the explanation below.

Work Step by Step

We use the trigonometric identity: $2\sin x\cos x=\sin 2x$. Now, the left side can be written as: $\begin{align} & \sin 2x\sec x=2\sin x\cos x\times \frac{1}{\cos x} \\ & =2\sin x \end{align}$ Thus, the left side is equal to $2\sin x$.
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