Simplifying $\frac{ \cos^2 x - \sin^2 x }{\sin{2x}}$ Please consider the following problem and my answer to it:
Problem:
Simplify the following expression:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} $$
Answer:
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} = \frac{ \cos^2 x - \sin^2 x }{ 2 \sin x \cos x} $$
$$ \frac{ \cos^2 x - \sin^2 x }{\sin{2x}} = \frac{\cos x}{2 \sin x} - \frac{\sin x}{2 \cos x} $$
The result is not much simpler than what I started with. Is there an additional simplification that can be made? Am I missing something?

Continue with 
$$ \frac{\cos x}{2 \sin x} - \frac{\sin x}{2 \cos x} 
=\frac1{2\tan x} -\frac12 \tan x
= \frac{1-\tan^2x}{2\tan x}=\frac1{\tan 2x}
$$
