A problem with trigonometric addition theorem

Question:
Show that $\frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\sin\alpha - \sin\beta}=\sin\alpha + \sin\beta$.

Answer:
Use $\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha$.
The given expression will become
\[ \frac{\sin(\alpha + \beta)\sin(\alpha - \beta)}{\sin\alpha - \sin\beta} = \frac{(\sin\alpha\cos\beta + \sin\beta\cos\alpha)(\sin\alpha\cos\beta - \sin\beta\cos\alpha)}{\sin\alpha - \sin\beta}
     = \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\sin\alpha - \sin\beta} \]

Use $\sin^2\theta+\cos^2\theta=1$ to eliminate $\cos^2\alpha$ and $\cos^2\beta$.
\[ \frac{\sin^2\alpha\cos^2\beta - \sin^2\beta\cos^2\alpha}{\sin\alpha - \sin\beta}
       = \frac{\sin^2\alpha(1-\sin^2\beta) - \sin^2\beta(1-\sin^2\alpha)}{\sin\alpha - \sin\beta}
      =  \frac{\sin^2\alpha-\sin^2\alpha\sin^2\beta - \sin^2\beta + \sin^2\beta\sin^2\alpha}{\sin\alpha - \sin\beta}
     = \frac{\sin^2\alpha - \sin^2\beta}{\sin\alpha - \sin\beta}
     = \frac{(\sin\alpha + \sin\beta)(\sin\alpha - \sin\beta)}{\sin\alpha - \sin\beta}
     = \sin\alpha + \sin\beta \]
QED

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