Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Test - Page 250: 2


$sec~\theta-sin~\theta~tan~\theta= cos~\theta$

Work Step by Step

We can express $sec~\theta-sin~\theta~tan~\theta$ as a single function of $\theta$: $sec~\theta-sin~\theta~tan~\theta$ $= \frac{1}{cos~\theta}-sin~\theta~\frac{sin~\theta}{cos~\theta}$ $= \frac{1}{cos~\theta}-\frac{sin^2~\theta}{cos~\theta}$ $= \frac{1-sin^2~\theta}{cos~\theta}$ $= \frac{cos^2~\theta}{cos~\theta}$ $= cos~\theta$
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.