Trigonometry (11th Edition) Clone

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

Chapter 5 - Trigonometric Identities - Section 5.2 Verifying Trigonometric Identities - 5.2 Exercises - Page 209: 71


$$\sin^2\alpha\sec^2\alpha+\sin^2\alpha\csc^2\alpha=\sec^2\alpha$$ The expression has been proved to be an identity.

Work Step by Step

$$\sin^2\alpha\sec^2\alpha+\sin^2\alpha\csc^2\alpha=\sec^2\alpha$$ The left is more complex and we need to deal with it first. $$A=\sin^2\alpha\sec^2\alpha+\sin^2\alpha\csc^2\alpha$$ - Using the Reciprocal Identities: $$\sec\alpha=\frac{1}{\cos\alpha}\hspace{1.5cm}\csc\alpha=\frac{1}{\sin\alpha}$$ Therefore $A$ would be $$A=\sin^2\alpha\times\frac{1}{\cos^2\alpha}+\sin^2\alpha\times\frac{1}{\sin^2\alpha}$$ $$A=\frac{\sin^2\alpha}{\cos^2\alpha}+1$$ Also, from Quotient Identity: $$\tan\alpha=\frac{\sin\alpha}{\cos\alpha}$$ Therefore, $$A=\tan^2\alpha+1$$ Finally, from Pythagorean Identity: $$\tan^2\alpha+1=\sec^2\alpha$$ So, $$A=\sec^2\alpha$$ The expression is hence an identity.
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