#### Answer

$\dfrac {1}{\pm \sqrt {1-\sin ^{2}x}}=\pm \dfrac {\sqrt {1-\sin ^{2}x}}{1-\sin ^{2}x}$

#### Work Step by Step

$secx=\dfrac {1}{\cos x}=\dfrac {1}{\pm \sqrt {1-\sin ^{2}x}}=\pm \dfrac {\sqrt {1-\sin ^{2}x}}{1-\sin ^{2}x}$

Published by
Pearson

ISBN 10:
013421742X

ISBN 13:
978-0-13421-742-0

$\dfrac {1}{\pm \sqrt {1-\sin ^{2}x}}=\pm \dfrac {\sqrt {1-\sin ^{2}x}}{1-\sin ^{2}x}$

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