Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 14 - Trigonometric Identities and Equations - 14-1 Trigonometric Identities - Practice and Problem-Solving Exercises: 28

Answer

$$\frac{1}{\sin\theta}$$

Work Step by Step

To simplify $$\frac{\tan\theta}{\sec\theta-\cos\theta},$$ substitute the Reciprocal Identity $$\sec\theta=\frac{1}{\cos\theta}$$ and the Tangent Identity $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ to obtain $$\frac{\tan\theta}{\sec\theta-\cos\theta}=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}-\cos\theta}.$$ Multiply both numerator and denominator by $\cos\theta$ $$=\frac{\frac{\sin\theta}{\cos\theta}}{\frac{1}{\cos\theta}-\cos\theta}\times \frac{\cos\theta}{\cos\theta}=\frac{\sin\theta}{1-\cos^{2}\theta}$$ Finally, use the Pythagorean Identity: $$\sin^{2}\theta+\cos^{2}\theta=1$$ rearranged as $$1-\cos^{2}\theta=\sin^{2}\theta$$ to substitute and obtain $$\frac{\tan\theta}{\sec\theta-\cos\theta}=\frac{\sin\theta}{\sin^{2}\theta}=\frac{1}{\sin\theta}$$
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