#### Answer

$\pm \sqrt {sec^{2}x-1}$

#### Work Step by Step

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

Published by
Pearson

ISBN 10:
013421742X

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

$\pm \sqrt {sec^{2}x-1}$

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