Since $\sec x = {1 \over {\cos x}} = (\cos x)^{ - 1} $, we use the chain rule and get
${{dy} \over {dx}} = ( - 1)(\cos x)^{ - 2} ( - \sin x)$
$ = {1 \over {\cos ^2 x}}\sin x$
$ = {1 \over {\cos x}}.{{\sin x} \over {\cos x}}$
$ = \sec x\tan x$