Answer
See graph, $x^2-y^2=1$
Work Step by Step
1. See graph for $(csc(t),cot(t))$ over $\frac{\pi}{4}\le t\le\frac{\pi}{2}$
2. Use the relation $csc^2t=cot^2t+1$, 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: 24
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