Trigonometry (11th Edition) Clone

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

Chapter 2 - Review Exercises - Page 94: 40



Work Step by Step

(Without a calculator) $60^{o}$ is a Special Angle, so from the table of Function Values of Special Angles, we find $\tan 60^{o}=\sqrt{3},\qquad \sec 60^{o}=2$ $LHS=1+(\sqrt{3})^{2}=1+3=4$ $RHS=(2)^{2}=4$ $LHS=RHS$ the statement is true
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