# 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.

• I think you need to divide your result that you worked out by 2 (as $\sin2x=2\sin x \cos x$) Commented Jan 10, 2021 at 18:12
• Write everything in terms of sin and cos. Then multiply through by all denominators to get a polynomial equation in sin and cos Commented Jan 10, 2021 at 18:15
• @A-LevelStudent Corrected. Commented Jan 10, 2021 at 18:22
• What is the question ? "How to simplify..."... is the problem to prove the identity? Or to simplify the equality and solve for all possible pairs $(\alpha, \beta)$ ? Commented Jan 10, 2021 at 18:28
• @AdamRubinson good point, I assumed we were proving an identity, as the question made no mention of solving an equation. If we are indeed solving an equation, then the question makes more sense, as the 'identity' is wrong. Commented Jan 10, 2021 at 18:30

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*}

• My bad. I happened to try out $2$ values that made the supposed answer only out by a factor of 2, hence my suggestions. Commented Jan 10, 2021 at 18:28
• @A-LevelStudent No worries, it happens. Commented Jan 10, 2021 at 18:28
• As noted by AdamRubinson the OP may in fact be attempting to solve an equation, not prove an identity. Commented Jan 10, 2021 at 18:30
• I will edit my answer then. Thanks for pointing out. Commented Jan 10, 2021 at 18:31
• No problem, I also made that mistake initially. Commented Jan 10, 2021 at 18:31