Answer
$\dfrac {\cos ^{4}x-\sin ^{4}x}{\cos ^{2}x}=1-\tan ^{2}x$
Work Step by Step
$\dfrac {\cos ^{4}x-\sin ^{4}x}{\cos ^{2}x}=\dfrac {\left( \cos ^{2}x-\sin ^{2}x\right) \left( \cos ^{2}x+\sin ^{2}x\right) }{\cos ^{2}x}=\dfrac {\cos ^{2}x-\sin ^{2}x}{\cos ^{2}x}=1-\tan ^{2}x$
