Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 5 - Trigonometric Functions - Section 5.2 Trigonometric Functions: Unit Circle Approach - 5.2 Assess Your Understanding - Page 402: 37

Answer

$\sqrt 6$

Work Step by Step

We need to compute $\csc 45^{\circ} \tan 60^{\circ}$. Note that: $\csc 45^{\circ}=\dfrac{1}{\sin 45^{\circ}}=\dfrac{1}{\frac{\sqrt2}{2}}=\sqrt2$ $\tan{60^{\circ}}=\sqrt3$ We the values above to obtain: $\csc 45^{\circ} \tan 60^{\circ} \\ = \sqrt2 \times \sqrt3 \\ =\sqrt 6$
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