Could someone please help me prove the following identity.

$\displaystyle \sec x + \tan x \equiv \dfrac{1}{\sec x -\tan x}$

I got as far as this starting with the R.H.S, but don't know if it's the right approach or how to proceed from here.

$\displaystyle R.H.S = \dfrac{\cos x}{1 - \sin x}$

Any help would be much appreciated, thank you.

Now solved. I multiplied the denominator and numerator by $\displaystyle 1 + \sin x$, then it was easy.