Help on proving a trigonometric identity involving cot and half angles Prove: $\cot\frac{x+y}{2}=-\left(\frac{\sin x-\sin y}{\cos x-\cos y}\right)$.
My original idea was to do this:
$\cot\frac{x+y}{2}$ = $\frac{\cos\frac{x+y}{2}}{\sin\frac{x+y}{2}}$, then substitute in the formulas for $\cos\frac{x+y}{2}$ and $\sin\frac{x+y}{2}$, but that became messy very quickly.
Did I have the correct original idea, but overthink it, or is there any easier way? Hints only, please.

Try using
$ \displaystyle \sin{x} - \sin{y} = \sin{\frac{2x}{2}} - \sin{\frac{2y}{2}} = 
2 \frac{\sin \left( {\frac{x+y}{2} + \frac{x-y}{2}} \right) - \sin \left( {\frac{x+y}{2} - \frac{x-y}{2}} \right)}{2}$
and then the product to sum-formula for sine, i.e.
$ \displaystyle \frac {\sin \left({x + y}\right) - \sin \left({x - y}\right)}{2} = \cos{x} \sin{y} $.
You can prove it geometrically as well.
