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: 39



Work Step by Step

$$\sec\theta=\csc\Big(\frac{\theta}{2}+20^\circ\Big)\hspace{1cm}(1)$$ We already get from the cofunction identity that $$\sec\theta=\csc(90^\circ-\theta)$$ That means using that in $(1)$, it can lead to $$\csc(90^\circ-\theta)=\csc\Big(\frac{\theta}{2}+20^\circ\Big)$$ $$90^\circ-\theta=\frac{\theta}{2}+20^\circ$$ $$\frac{3\theta}{2}=70^\circ$$ $$\theta=\frac{70^\circ}{\frac{3}{2}}=\frac{70^\circ\times2}{3}$$ $$\theta=\frac{140^\circ}{3}$$
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.