Answer
See graph, $x^2-y^2=1$
Work Step by Step
1. See graph for $(sec(t),tan(t))$ over $0\le t\le\frac{\pi}{4}$
2. Use the relation $sec^2t=tan^2t+1$, we have $x^2=y^2+1$ or $x^2-y^2=1$
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 9 - Analytic Geometry - Section 9.7 Plane Curves and Parametric Equations - 9.7 Assess Your Understanding - Page 714: 23
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
