Precalculus (6th Edition) Blitzer

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

Chapter 4 - Review Exercises - Page 644: 35



Work Step by Step

The trigonometric ratios are as follows: $\sin \theta= \dfrac{opposite}{hypotenuse}$ ; $\cos \theta= \dfrac{Adjacent}{hypotenuse}$ and $\tan \theta= \dfrac{Opposite}{Adjacent}$ Now, $\sec \theta= \dfrac{1}{\dfrac{Adjacent}{hypotenuse}}$ So, $\cos (\dfrac{2 \pi}{9}) \times \sec (\dfrac{2 \pi}{9})=\cos (\dfrac{2 \pi}{9}) \times \dfrac{1}{\cos (\dfrac{2 \pi}{9})}$ Our required answer is: $1$
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