Solved :secx + tanx = 1/(secx-tanx)

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.

Re: secx + tanx = 1/(secx-tanx)

You are correct, now just write it as 1/[(1-sinx)/cosx] and proceed further.

Alternatively multiply and divide the numerator and denominator by ( secx-tanx) the numerator will reduce to 1 and you will get the RHS.

Re: secx + tanx = 1/(secx-tanx)

Thank you very much for your reply, sorry I did not say in my original post, but I actually started with the R.H.S. I've now edited my post accordingly thank you.