Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 2 - Acute Angles and Right Triangles - Section 2.2 Trigonometric Functions of Non-Acute Angles - 2.2 Exercises - Page 62: 50

Answer

$\cot$$^{2}$ 90$^{\circ}$ - $\sec$$^{2}$ 180$^{\circ}$ + $\csc$$^{2}$ 135$^{\circ}$ = 1

Work Step by Step

$\cot$$^{2}$ 90$^{\circ}$ - $\sec$$^{2}$ 180$^{\circ}$ + $\csc$$^{2}$ 135$^{\circ}$ $\cot$ 90$^{\circ}$ = 0 $\sec$ 180$^{\circ}$ = -1 $\csc$ 135$^{\circ}$ = $\csc$ 135$^{\circ}$(In Quadrant II) =$\sqrt2$ Therefore: = 0$^{2}$ - (-1)$^{2}$ + ($\sqrt2$)$^{2}$ = -1 + 2 = 1 $\cot$$^{2}$ 90$^{\circ}$ - $\sec$$^{2}$ 180$^{\circ}$ + $\csc$$^{2}$ 135$^{\circ}$ = 1
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