Prove: $\sin (\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha$ 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?
 A: This website has a lot of cool stuff about trigonometry.
https://trigonography.com/2015/09/28/angle-sum-and-difference-for-sine-and-cosine/

A: 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
$$ \sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta. $$
