This is an exercise in Gelfand's Trigonometry, It is not that difficult but I am doing something wrong that is preventing me from proving the identity.
We need to use the following diagram to prove it: 
My attempt:
$$ \begin{eqnarray*} \sin (\alpha - \beta) = \frac{CD}{AC} \\ = \frac{PQ}{AC} \\ = \frac{BQ - BP}{AC} \\ = \frac{BQ}{AC} - \frac{BP}{AC} \\ \end{eqnarray*} $$
Now in the following step we should use an intermediary to make this equal to the required identity, but for the first fraction I can't find anything rather than $AB$} $$ = \frac{BQ}{AB} \cdot \frac{AB}{AC} \\ $$ My problem here is I don't see how $\frac{AB}{AC}$ would simplify to $\cos \beta$ to me this seems like $\sec \beta$ How could this be fixed?
I don't know if you like this or not. Let $AC=1$. Then in $rt\Delta ACD$, $$ \sin(\alpha-\beta)=CD=PQ=BQ-BP.$$ In $rt\Delta ABC$, $AC=AB\cos\beta$ and hence $AB=\frac1{\cos\beta}, BC=\tan\beta.$ So in $rt\Delta ABQ$, $$BQ=AB\sin\alpha=\frac{\sin\alpha}{\cos\beta}.$$ Also in $rt\Delta BPC$, $\angle PBC=\alpha-\beta$ and hence $$ BP=BC\cos(\alpha-\beta)=\tan\beta\cos(\alpha-\beta). $$ So one has $$ \sin(\alpha-\beta)=\frac{\sin\alpha}{\cos\beta}-\tan\beta\cos(\alpha-\beta). \tag{1}$$ Similarly $$ \cos(\alpha-\beta)=\frac{\cos\alpha}{\cos\beta}+\tan\beta\sin(\alpha-\beta). \tag{2}$$ Putting (2) in (1), one has \begin{eqnarray} \sin(\alpha-\beta)&=&\frac{\sin\alpha}{\cos\beta}-\tan\beta\cos(\alpha-\beta)\\ &=&\frac{\sin\alpha}{\cos\beta}-\tan\beta\left(\frac{\cos\alpha}{\cos\beta}+\tan\beta\sin(\alpha-\beta)\right)\\ &=&\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos^2\beta}-\tan^2\beta\sin(\alpha-\beta) \end{eqnarray} or $$(1+\tan^2\beta)\sin(\alpha-\beta)=\frac{\sin\alpha\cos\beta-\cos\alpha\sin\beta}{\cos^2\beta} $$ or $$ \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta. $$