## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 4 - Section 4.3 - Right Triangle Trigonometry - Exercise Set - Page 561: 23

#### Answer

$\sec{65^o}$

#### Work Step by Step

RECALL: The cofunction Identities: (1) $\sin{\theta} = \cos{(90^o-\theta)}$ (2) $\cos{\theta} = \sin{(90^o-\theta)}$ (3) $\tan{\theta} = \cot{(90^o-\theta)}$ (4) $\cot{\theta} = \tan{(90-\theta^o)}$ (5) $\csc{\theta} = \sec{(90^o-\theta)}$ (6) $\sec{\theta} = \csc{(90^0-\theta)}$ Use identity (5) to obtain: $\csc{25^o} = \sec{(90^o-25^o)} = \sec{65^o}$

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