Precalculus (6th Edition)

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

Chapter 7 - Trigonometric Identities and Equations - 7.2 Verifying Trigonometric Identities - 7.2 Exercises - Page 668: 84

Answer

$$\frac{{1 - \cos \theta }}{{1 + \cos \theta }} = 2{\csc ^2}\theta - 2\csc \theta \cot \theta - 1$$

Work Step by Step

$$\eqalign{ & \frac{{1 - \cos \theta }}{{1 + \cos \theta }} = 2{\csc ^2}\theta - 2\csc \theta \cot \theta - 1 \cr & {\text{We transform the more complicated left side to match the right side}}. \cr & \frac{{1 - \cos \theta }}{{1 + \cos \theta }} = \frac{{1 - \cos \theta }}{{1 + \cos \theta }} \cdot \frac{{1 - \cos \theta }}{{1 - \cos \theta }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{1 - {{\cos }^2}\theta }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1 - 2\cos \theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{{{{\sin }^2}\theta }} - \frac{{2\cos \theta }}{{{{\sin }^2}\theta }} + \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\csc ^2}\theta - 2\left( {\frac{1}{{\sin \theta }}} \right)\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right) + {\cot ^2}\theta \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\csc ^2}\theta - 2\csc \theta \cot \theta + \left( {{{\csc }^2}\theta - 1} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2{\csc ^2}\theta - 2\csc \theta \cot \theta - 1 \cr & {\text{Thus have verified that the given equation is an identit}} \cr} $$
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