Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Mid-Chapter Check Point - Page 577: 15


$\sqrt {35}$

Work Step by Step

Since, $\sec^2 \theta=1+\tan^2 \theta $ $\implies \tan \theta=\sqrt {1+\dfrac{1}{\cos^2 \theta}}$ or, $36=1+a^2 \implies a=\sqrt {35}$ Thus, $\cos (\dfrac{\pi}{2}- \theta)=\tan \theta=\sqrt {35}$
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