Trigonometry (11th Edition) Clone

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

Chapter 5 - Test - Page 250: 10


$sec^2~B = \frac{1}{cos^2~B} = \frac{1}{1-sin^2~B}$

Work Step by Step

We can verify the identity: $sec^2~B = \frac{1}{cos^2~B} = \frac{1}{1-sin^2~B}$
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