Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Review Exercises - Page 248: 12



Work Step by Step

$$A=\csc^2\theta+\sec^2\theta$$ $\csc\theta$ and $\sec\theta$ can be written according to the following identities: $$\csc\theta=\frac{1}{\sin\theta}\hspace{2cm}\sec\theta=\frac{1}{\cos\theta}$$ So, $$A=\frac{1}{\sin^2\theta}+\frac{1}{\cos^2\theta}$$ $$A=\frac{\cos^2\theta+\sin^2\theta}{\sin^2\theta\cos^2\theta}$$ - Recall the identity $\cos^2\theta+\sin^2\theta=1$ $$A=\frac{1}{\sin^2\theta\cos^2\theta}$$ This is the answer we need to the question.
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