Prove the identity:
$\dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right) = \dfrac{\sin \beta}{\sin \alpha}$
LHS $= \dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right)$
$= \dfrac{\sin \left[\left(\alpha + \alpha\right) + \beta\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha }$
$= \dfrac{\sin \left[\alpha + \left(\alpha + \beta\right)\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$
$= \dfrac{\sin \alpha \cos \left(\alpha + \beta\right) + \cos \alpha \sin \left(\alpha + \beta\right) - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$
$= \dfrac{\sin \left(\alpha + \beta\right) \cos \alpha - \cos \left(\alpha + \beta\right) \sin \alpha}{\sin \alpha}$
$= \dfrac{\sin \left[\left(\alpha + \beta\right) - \alpha\right]}{\sin \alpha}$
$= \dfrac{\sin \beta}{\sin \alpha} = $ RHS