[tex]\sec x-\tan x=\dfrac{1}{\sec x+\tan x}\\\\
\dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}=\dfrac{1}{\frac{1}{\cos x}+\frac{\sin x}{\cos x}}\\\\
\dfrac{1-\sin x}{\cos x}=\dfrac{1}{~~\frac{1+\sin x}{\cos x}~~}\\\\\\
(1-\sin x)(\dfrac{1+\sin x}{\cos x})=\cos x\\\\
\dfrac{1-\sin^2x}{\cos x}=\cos x\\\\
\dfrac{\cos^2x}{\cos x}=\cos x\\\\
\cos x=\cos x\Longrightarrow OK!~~\blacksquare[/tex]
