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?

$\cos^2x -\sin^2 x = \cos (2x)$. So the result can be written as $\cot(2x)$.
