Simplify a trigonometric expression I've been simplying a slew of trigonometric expressions and most of them fall out pretty clearly, but this one has been giving me fits:
$$\frac{(\sec x - \tan x)^{2} + 1}{\sec x\csc x - \tan x\csc x}$$
Here's my best attempt so far:
$$\begin{align*}
&=\frac{\left(\frac{1}{\cos x}-\frac{\sin x}{\cos x} \right)^{2} + 1}{\frac{1}{\cos x\sin x}-\frac{1}{\cos x}}\\\\
&=\frac{\frac{1}{\cos^{2}x}-2\frac{\sin x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} + 1}{\frac{\cos x-\cos x\sin x}{\cos^{2}x\sin x}}
\end{align*}$$
But I can't help but think I've already gone wrong since I feel like I should start factoring things out again. Am I on the right track here?
Thanks for any suggestions.

You have
$$\begin{align*}
&=\frac{(\frac{1}{\cos x}-\frac{\sin x}{\cos x})^{2} + 1}{\frac{1}{\cos x\sin x}-\frac{1}{\cos x}}\\
&=\frac{\frac{1}{\cos^{2}x}-2\frac{\sin x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x} + 1}{\frac{\cos x-\cos x\sin x}{\cos^{2}x\sin x}}
\end{align*}$$
Your next steps might be to note that on top, $\sin x/\cos x = \tan x$ and $1 + \tan^2 x= \sec^2 x$, leaving you with $2\dfrac{(1 -  \sin x)}{\cos ^2 x}$ 
On the bottom, you have $\dfrac{\cos x}{\sin x}\dfrac{(1 - \sin x)}{\cos ^2 x}$
So cancel all that you can, and I think you'll be able to finish from here.
