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 5.3 Sum and Difference Identities for Cosine - 5.3 Exercises - Page 219: 42

Answer

$$\theta=40^\circ$$

Work Step by Step

$$\cot(\theta-10^\circ)=\tan(2\theta-20^\circ)\hspace{1cm}(1)$$ Cofunction identity: $$\cot\theta=\tan(90^\circ-\theta)$$ So, $$\cot(\theta-10^\circ)=\tan[90^\circ-(\theta-10^\circ)]$$ $$\cot(\theta-10^\circ)=\tan(90^\circ-\theta+10^\circ)$$ $$\cot(\theta-10^\circ)=\tan(100^\circ-\theta)$$ Replace this in $(1)$ for $\cot(\theta-10^\circ)$: $$\tan(100^\circ-\theta)=\tan(2\theta-20^\circ)$$ $$100^\circ-\theta=2\theta-20^\circ$$ $$3\theta=120^\circ$$ $$\theta=40^\circ$$
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