Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises - Page 59: 58

Answer

True

Work Step by Step

$\tan$$^{2}$ 60$^{\circ}$ + 1 = $\sec$$^{2}$ 60$^{\circ}$ First evaluate the value of each trigonometric statement: $\tan$ 60$^{\circ}$ = $\sqrt3$ $\sec$ 60$^{\circ}$ = 2 Next evaluate each side of the equation: $\tan$$^{2}$ 60$^{\circ}$ + 1 = ($\sqrt3$)$^{2}$ + 1 = 3 + 1 = 4 $\sec$$^{2}$ 60$^{\circ}$ = 2$^{2}$ = 4 Therefore: $\tan$$^{2}$ 60$^{\circ}$ + 1 = $\sec$$^{2}$ 60$^{\circ}$ is a true statement.
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.