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.
Simplify $(\frac{\sin\alpha – \sin\beta}{\cos\beta -\cos\alpha})\cdot \cos\alpha \cdot \cos\beta$
trigonometry
Related Solutions
This website has a lot of cool stuff about trigonometry.
https://trigonography.com/2015/09/28/angle-sum-and-difference-for-sine-and-cosine/
$$\sin\alpha+\sin\beta=2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=a$$ And $$\cos\alpha+\cos\beta=2\cos\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=b$$
Dividing, $$\tan\left(\frac{\alpha+\beta}{2}\right)=\frac ab=t$$
However, $$\sin(\alpha+\beta)=\frac{2t}{1+t^2}$$ and $$\cos(\alpha+\beta)=\frac{1-t^2}{1+t^2}$$
And it is readily seen that both results follow directly from this.
Best Answer
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*}