1. Datos y fórmulas:
Utilizaremos $\tan(\beta - \alpha) = \frac{\tan \beta - \tan \alpha}{1 + \tan \beta \tan \alpha}$.

2. Desarrollo:
Sustituimos $\tan \alpha$ en la fórmula:
$$ \tan(\beta - \alpha) = \frac{\tan \beta - \frac{Q \sin \beta}{P + Q \cos \beta}}{1 + \tan \beta \cdot \frac{Q \sin \beta}{P + Q \cos \beta}} $$
Escribimos $\tan \beta = \frac{\sin \beta}{\cos \beta}$:
$$ \tan(\beta - \alpha) = \frac{\frac{\sin \beta}{\cos \beta} - \frac{Q \sin \beta}{P + Q \cos \beta}}{1 + \frac{\sin \beta}{\cos \beta} \cdot \frac{Q \sin \beta}{P + Q \cos \beta}} $$

Operamos el numerador:
$$ \frac{\sin \beta (P + Q \cos \beta) - Q \sin \beta \cos \beta}{\cos \beta (P + Q \cos \beta)} = \frac{P \sin \beta + Q \sin \beta \cos \beta - Q \sin \beta \cos \beta}{\cos \beta (P + Q \cos \beta)} = \frac{P \sin \beta}{\cos \beta (P + Q \cos \beta)} $$

Operamos el denominador:
$$ \frac{\cos \beta (P + Q \cos \beta) + Q \sin^2 \beta}{\cos \beta (P + Q \cos \beta)} = \frac{P \cos \beta + Q \cos^2 \beta + Q \sin^2 \beta}{\cos \beta (P + Q \cos \beta)} = \frac{P \cos \beta + Q(1)}{\cos \beta (P + Q \cos \beta)} $$

Dividiendo los resultados:
$$ \tan(\beta - \alpha) = \frac{P \sin \beta}{Q + P \cos \beta} $$

3. Conclusión:
$$ \boxed{\tan(\beta - \alpha) = \frac{P \sin \beta}{Q + P \cos \beta}} $$