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: 16

Answer

$${\coth ^2}x - 1 = {\operatorname{csch} ^2}x$$

Work Step by Step

$$\eqalign{ & {\cosh ^2}x - {\sinh ^2}x = 1 \cr & {\text{divide by sin}}{{\text{h}}^2}x \cr & \frac{{{{\cosh }^2}x}}{{{\text{sin}}{{\text{h}}^2}x}} - \frac{{{{\sinh }^2}x}}{{{\text{sin}}{{\text{h}}^2}x}} = \frac{1}{{{\text{sin}}{{\text{h}}^2}x}} \cr & {\text{simplify}} \cr & \frac{{{{\cosh }^2}x}}{{{\text{sin}}{{\text{h}}^2}x}} - 1 = \frac{1}{{{\text{sin}}{{\text{h}}^2}x}} \cr & {\text{use hyperbolic functions cosecant and cotangent}} \cr & {\coth ^2}x - 1 = {\operatorname{csch} ^2}x \cr} $$
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