Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises: 14

Answer

$$\sin x\tan x = \sin x\tan x$$

Work Step by Step

$$\eqalign{ & 2\sinh \left( {\ln \left( {\sec x} \right)} \right) = \sin x\tan x \cr & {\text{definition of }}sinhx \cr & 2\left( {\frac{{{e^{\ln \sec x}} - {e^{ - \ln \sec x}}}}{2}} \right) = \sin x\tan x \cr & {\text{simplify}} \cr & {e^{\ln \sec x}} - {e^{ - \ln \sec x}} = \sin x\tan x \cr & \sec x - \frac{1}{{\sec x}} = \sin x\tan x \cr & {\text{reciprocal identity}} \cr & \frac{1}{{\cos x}} - \cos x = \sin x\tan x \cr & \frac{{1 - {{\cos }^2}x}}{{\cos x}} = \sin x\tan x \cr & {\text{identity si}}{{\text{n}}^2}x + {\cos ^2}x = 1 \cr & \frac{{{{\sin }^2}x}}{{\cos x}} = \sin x\tan x \cr & \sin x\frac{{\sin x}}{{\cos x}} = \sin x\tan x \cr & \sin x\tan x = \sin x\tan x \cr} $$
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