Precalculus (6th Edition) Blitzer

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

Chapter 5 - Review Exercises - Page 707: 2

Answer

The left side is equal to the right side

Work Step by Step

The expression on the left side $\text{cos }x+\sin x.\tan x$ can be simplified by using the quotient identity $\tan x=\frac{\sin x}{\cos x}$ $\begin{align} & \text{cos }x+\sin x.\tan x=\frac{\cos x}{\cos x}.\cos x+\sin x.\frac{\sin x}{\cos x} \\ & =\frac{{{\cos }^{2}}x}{\cos x}+\frac{{{\sin }^{2}}x}{\cos x} \\ & =\frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\cos x} \end{align}$ Use the Pythagorean identity, ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Therefore, the expression can be further simplified as: $\frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\cos x}=\frac{1}{\cos x}$ As per the reciprocal identity $\frac{1}{\cos x}=\sec x$ $\begin{align} & \frac{{{\cos }^{2}}x+{{\sin }^{2}}x}{\cos x}=\frac{1}{\cos x} \\ & =\sec x \end{align}$ Thus, the left side is equal to the right side $\text{cos }x+\sin x.\tan x=\sec x$.
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