Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 6 - Analytic Trigonometry - Section 6.4 Trigonometric Identities - 6.4 Assess Your Understanding - Page 497: 47


See proof below.

Work Step by Step

We know that $\sec \theta =\dfrac{1}{\cos \theta}$ and $\csc \theta=\dfrac{1}{\sin \theta}$ In order to prove the given identity, we simplify the left hand side $\text{LHS}$ as follows: $\text{LHS } =\dfrac{\sec \theta}{\csc \theta} +\dfrac{\sin \theta}{\cos \theta} \\= \dfrac{1/\cos \theta}{1/\sin \theta} +\dfrac{\sin \theta}{\cos \theta} \\ =\dfrac{\sin }{\cos \theta} +\dfrac{\sin \theta}{\cos \theta} \\=\tan \theta + \tan \theta \\= 2 \tan \theta \\= \text{ RHS}$
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