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