Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapters 1-3 - Cumulative Test - Page 176: 10

Answer

Please refer to the step-by-step part for the proof.

Work Step by Step

RECALL: (1) $\csc{\theta}=\dfrac{1}{\sin{\theta}}$ (2) $\cot{\theta} = \dfrac{\cos{\theta}}{\sin{\theta}}$ Work on the left hand side of the equation to obtain: $\require{cancel} \sin{\theta}(\csc{\theta} + \cot{\theta}) \\=\sin{\theta}\left(\dfrac{1}{\sin{\theta}} + \dfrac{\cos{\theta}}{\sin{\theta}}\right) \\=\sin{\theta}\cdot \dfrac{1}{\sin{\theta}} + \sin{\theta} \cdot \dfrac{\cos{\theta}}{\sin{\theta}} \\=\cancel{\sin{\theta}}\cdot \dfrac{1}{\cancel{\sin{\theta}}} + \cancel{\sin{\theta}} \cdot \dfrac{\cos{\theta}}{\cancel{\sin{\theta}}} \\=1+\cos{\theta}$ The two sides of the given equation was shown to be equal, therefore $\sin{\theta}(\csc{\theta} + \cot{\theta}) = 1+\cos{\theta}$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.