Precalculus (6th Edition) Blitzer

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

Chapter 5 - Section 5.1 - Verifying Trigonometric Identities - Exercise Set - Page 659: 55

Answer

See the explanation below.

Work Step by Step

$\left( {{\tan }^{2}}\theta +1 \right)\left( {{\cos }^{2}}\theta +1 \right)$ We multiply the above expression as: $\left( {{\tan }^{2}}\theta +1 \right)\left( {{\cos }^{2}}\theta +1 \right)={{\tan }^{2}}\theta {{\cos }^{2}}\theta +{{\tan }^{2}}\theta +{{\cos }^{2}}\theta +1$ By using the quotient identity of trigonometry ${{\tan }^{2}}\theta =\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }$ , the above expression can be further simplified as: $\begin{align} & {{\tan }^{2}}\theta {{\cos }^{2}}\theta +{{\tan }^{2}}\theta +{{\cos }^{2}}\theta +1=\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }.{{\cos }^{2}}\theta +{{\tan }^{2}}\theta +{{\cos }^{2}}\theta +1 \\ & ={{\sin }^{2}}\theta +{{\tan }^{2}}\theta +{{\cos }^{2}}\theta +1 \\ & ={{\sin }^{2}}\theta +{{\cos }^{2}}\theta +{{\tan }^{2}}\theta +1 \end{align}$ Apply the Pythagorean identity of trigonometry ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$; now, the above expression can be further simplified as: $\begin{align} & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +{{\tan }^{2}}\theta +1=1+{{\tan }^{2}}\theta +1 \\ & ={{\tan }^{2}}\theta +2 \end{align}$ Thus, the left side of the expression is equal to the right side, which is $\left( {{\tan }^{2}}\theta +1 \right)\left( {{\cos }^{2}}\theta +1 \right)={{\tan }^{2}}\theta +2$.
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