Proving that $\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2}$ How would I prove the following trig identity? 


  $$\frac{ \cos (A+B)}{ \cos A-\cos B}=-\cot \frac{A-B}{2} \cot \frac{A+B}{2} $$


My work thus far has been:
$$\dfrac{2\cos\dfrac{A+B}{2} \cos\dfrac{A-B}{2}}{-2\sin\dfrac{A+B}{2} \sin\dfrac{A-B}{2}} =-\cot\dfrac{A+B}{2} \cot\dfrac{A-B}{2} \ .$$
First, note that $\dfrac{\cos(A + B)}{\cos A - \cos B} \neq -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$

In particular, if we use something like $A = \pi/6, B = 2\pi/6$, then the left is $0$ as $\cos(\pi/2) = 0$ and the right side is a product of two nonzero things.

I suspect instead that you would like to prove:

$$\dfrac{\cos A + \cos B}{\cos A - \cos B} = -\cot \left( \dfrac{A-B}{2} \right) \cot \left(\dfrac{A+B}{2} \right)$$

HINTS

And this looks to me like an exercise in the sum-to-product and product-to-sum trigonometric identities (wiki reference). In fact, if you just apply these identities to the top and the bottom, you'll get the result.
A couple of hints: use the $\cot$ identity on the RHS to put in terms of $\sin$ and $\cos$, use a product to sum identity, and pay attention to odd and even functions. This should simplify it to an easier expression.