Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 6 - Trigonometric Functions - 6.3 Properties of the Trigonometric Functions - 6.3 Assess Your Understanding - Page 395: 84



Work Step by Step

I know that $\tan$ has a period of $180^\circ$, hence $\tan{\theta}=\tan{(\theta-180^\circ)}.$ Hence: $\tan{200^\circ}\cot{20^\circ}=\tan{(200^\circ-180^\circ)}\cot{20^\circ}=\tan{20^\circ}\cot{20^\circ}$. We also know that $\tan{\theta}\cot{\theta}=1$. Thus: $\tan{20^\circ}\cot{20^\circ}=1$.
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