Can someone help me simplify this trig expression? 
$$( \tan x+ \sec x )( \cot x-\cos x ) $$

I got stuck after a few steps of converting and adding and what not.
 A: Let $a = \tan x$ and $b = \sec x$; then the expression is equal to
$$\left(a + b\right)\left(\frac 1 a - \frac 1 b\right) = 1 + \frac b a - \frac a b - 1 = \frac b a - \frac a b$$
Now note that $\frac 1 a = \frac{\cos x}{\sin x}$ and $\frac 1 b = \cos x$ to finish the simplification.
A: A simplification:
$$
\begin{align*}
\left( \tan x + \sec x \right) \left( \cot x - \cos x  \right) &= \left(\tan x + \frac{1}{\cos x} \right)\left( \frac{1}{\tan x } - \cos x\right) \\
&=\frac{\tan x}{\tan x}-\cos x \tan x +\frac{1}{\cos x \tan x}-\frac{\cos x}{\cos x} \\
&=1-\cos x \tan x +\frac{1}{\cos x \tan x}-1 \\
&= \frac{1}{\cos x \tan x}-\cos x \tan x \\
&= \frac{\cos x}{\cos x \sin x}-\frac{\cos x \sin x}{\cos x} \\
&= \frac{1}{\sin x}-\sin x \\
&= \csc x - \sin x.
\end{align*}
$$
A: $$( \tan x+ \sec x )( \cot x-\cos x )=\frac{1+\sin x}{\cos x}\frac{\cos x(1-\sin x)}{\sin x} $$
if $\cos x\ne0,$ this becomes $$\frac{(1+\sin x)(1-\sin x)}{\sin x}=\frac{1-\sin^2x}{\sin x}=\frac{\cos^2x}{\sin x}$$
I was expecting something more interesting :) 
