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