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: 30


See the explanation below.

Work Step by Step

$1-\frac{{{\cos }^{2}}x}{1+\sin x}=\sin x$ Recall Pythagorean Identity, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ Use the above identity and solve the left side of the given expression, $1-\frac{{{\cos }^{2}}x}{1+\sin x}=\sin x$ Multiply numerator and denominator of $\frac{{{\cos }^{2}}x}{1+\sin x}$ by $1-\sin x$. $\begin{align} & 1-\frac{{{\cos }^{2}}x}{1+\sin x}=1-\frac{{{\cos }^{2}}x}{1+\sin x}\cdot \left( \frac{1-\sin x}{1-\sin x} \right) \\ & =1-\frac{{{\cos }^{2}}x\left( 1-\sin x \right)}{1-{{\sin }^{2}}x} \\ & =1-\frac{{{\cos }^{2}}x\left( 1-\sin x \right)}{{{\cos }^{2}}x} \\ & =1-\left( 1-\sin x \right) \end{align}$ Further simplify, $\begin{align} & 1-\frac{{{\cos }^{2}}x}{1+\sin x}=1-1+\sin x \\ & =\sin x \end{align}$ Therefore, $1-\frac{{{\cos }^{2}}x}{1+\sin x}=\sin x$
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