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: 51

Answer

$\cos$$^{2}$ 60$^{\circ}$ + $\sec$$^{2}$ 150$^{\circ}$ - $\csc$$^{2}$ 210$^{\circ}$ = -$\frac{29}{12}$

Work Step by Step

$\cos$$^{2}$ 60$^{\circ}$ + $\sec$$^{2}$ 150$^{\circ}$ - $\csc$$^{2}$ 210$^{\circ}$ $\cos$ 60$^{\circ}$ = $\frac{1}{2}$ $\sec$ 150$^{\circ}$ = -$\frac{2}{\sqrt3}$ $\csc$ 210$^{\circ}$ = $\csc$ 30$^{\circ}$(In Quadrant III) = -2 Therefore: = ($\frac{1}{2}$)$^{2}$ - (-$\frac{2}{\sqrt3}$)$^{2}$ - (-2)$^{2}$ = $\frac{1}{4}$ + $\frac{4}{3}$ - 4 = $\frac{3+16-48}{12}$ = -$\frac{29}{12}$ $\cos$$^{2}$ 60$^{\circ}$ + $\sec$$^{2}$ 150$^{\circ}$ - $\csc$$^{2}$ 210$^{\circ}$ = -$\frac{29}{12}$
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