trig proof help please, my work is attached I worked on both sides and thought i would end up with the original equation? my work is attached.

 A: $$\frac{ \sec x }{ \sec x -\tan x }=\frac{ \sec x }{ \sec x -\tan x }\times\frac{ \sec x +\tan x}{ \sec x +\tan x } =\frac{ \sec x(\sec x+ \tan x) }{ \sec ^2x -\tan^2 x }=\frac{ \sec x(\sec x+ \tan x) }{ 1 }$$
A: On your third line you lost parentheses. 
$\dfrac{\sec x}{\sec x - \tan x} = \sec^2 x + \sec x \tan x$
$ \iff \dfrac{\sec x}{\sec x - \tan x} = \sec x[\sec x + \tan x] 
$
$\iff \dfrac{1}{\sec x - \tan x} = \sec x + \tan x $
$\iff 1 = \sec^2 x - \tan^2 x $
A: On the second line you worked on both sides of the equation. After that you only worked on the right-hand side.
So, you could just work on the right-hand side completely and end up like the left-hand side. So, on step 2 instead of working on both sides, just multiply the right side by (secx - tanx) and also multiply the right side by one over (secx - tanx). Then do exactly what you have done and at the last step multiply what you have done times the factor one over (secx - tanx) which has been waiting all this time on the right of the right-hand side.
Convert the 1/cosx  to the equivilent secx and multiply times the 1/(secx - tanx) to get the same as the left-hand side.
