Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 5 - Trigonometric Identities - Section 5.1 Fundamental Identities - 5.1 Exercises - Page 196: 79

Answer

$$\frac{1+\tan^2\theta}{1+\cot^2\theta}=\tan^2\theta$$

Work Step by Step

$$A=\frac{1+\tan^2\theta}{1+\cot^2\theta}$$ - Quotient Identities: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ $$\cot\theta=\frac{\cos\theta}{\sin\theta}$$ Replace into $A$: $$A=\frac{1+\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\cos^2\theta}{\sin^2\theta}}$$ $$A=\frac{\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}}{\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}}$$ $$A=\frac{(\cos^2\theta+\sin^2\theta)\sin^2\theta}{\cos^2\theta(\cos^2\theta+\sin^2\theta)}$$ $$A=\frac{\sin^2\theta}{\cos^2\theta}$$ $$A=\Bigg(\frac{\sin\theta}{\cos\theta}\Bigg)^2$$ - Quotient Identity again: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ Replace into $A$: $$A=\tan^2\theta$$
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