Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 7 - Section 7.1 - Trigonometric Identities - 7.1 Exercises - Page 543: 27

Answer

$ 2 \sec^{2} α $

Work Step by Step

Given expression is- $ \frac{1}{1 - \sin α} + \frac{1}{1 + \sin α} $ = $ \frac{1 + \sin α + 1 - \sin α}{(1 - \sin α)(1 + \sin α)} $ = $ \frac{2}{(1 - \sin^{2} α)} $ = $ 2 . \frac{1}{(\cos^{2} α)} $ ( From first Pythagorean identity, $1 - \sin^{2} α$ = $\cos^{2} α$) = $ 2 \sec^{2} α $
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