<s>[INST] How to simplify $(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha})\cdot \cos\alpha \cdot \cos\beta$? There is this problem:
$$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{1}{\tan\alpha -\tan\beta}$$
I started as $$\left(\frac{\sin\alpha - \sin\beta}{\cos\beta -\cos\alpha}\right)\cdot \cos\alpha \cdot \cos\beta = \frac{\sin2\alpha \cos\beta-\sin2\beta \cos\alpha}{2(\cos\beta -\cos\alpha)}$$ but I'm stuck here, because I don't see how could I use $\sin(x-y)$ but also don't see how could I use any other identity without complicating this even more.
 [/INST] The proposed identity is not true (even though after the correction).
Indeed, take $\alpha = \pi/4$ and $\beta = 0$. Then we obtain:
\begin{align*}
\left(\frac{\sin(\pi/4) - \sin(0)}{\cos(0) - \cos(\pi/4)}\right)\times\cos(\pi/4)\cos(0) = \frac{\sqrt{2}}{2 - \sqrt{2}}\times\frac{\sqrt{2}}{2} = \frac{1}{2-\sqrt{2}}
\end{align*}
On the other hand, one has that
\begin{align*}
\frac{1}{2(\tan(\pi/4) - \tan(0))} = \frac{1}{2}
\end{align*}
EDIT
The proposed equation is equivalent to
\begin{align*}
\frac{(\sin(a) - \sin(b))\sin(a-b)}{\cos(b) - \cos(a)} = 1
\end{align*}
</s>