1. Datos del problema:

• $A + B + C = 180^\circ$.
• Expresión: $M = \frac{\sin 2A + \sin 2B - \sin 2C}{\sin 2A - \sin 2B + \sin 2C}$.

2. Fórmulas/Propiedades:

• $\sin X + \sin Y = 2 \sin\left(\frac{X+Y}{2}\right) \cos\left(\frac{X-Y}{2}\right)$.
• $\sin 2\theta = 2 \sin \theta \cos \theta$.
• $\sin(A+B) = \sin C$ y $\cos(A+B) = -\cos C$ (cuando $A+B+C=180^\circ $).

3. Desarrollo paso a paso:
Simplificamos el numerador ($ N $):
$$ N = (\sin 2A + \sin 2B) - \sin 2C = 2 \sin(A+B) \cos(A-B) - 2 \sin C \cos C $$
$$ N = 2 \sin C \cos(A-B) - 2 \sin C \cos C = 2 \sin C [\cos(A-B) - \cos C] $$
Como $\cos C = -\cos(A+B)$:
$$ N = 2 \sin C [\cos(A-B) + \cos(A+B)] = 2 \sin C [2 \cos A \cos B] = 4 \sin C \cos A \cos B $$

Simplificamos el denominador ($D$):
$$ D = (\sin 2A + \sin 2C) - \sin 2B = 2 \sin(A+C) \cos(A-C) - 2 \sin B \cos B $$
$$ D = 2 \sin B \cos(A-C) - 2 \sin B \cos B = 2 \sin B [\cos(A-C) - \cos B] $$
Como $\cos B = -\cos(A+C)$:
$$ D = 2 \sin B [\cos(A-C) + \cos(A+C)] = 2 \sin B [2 \cos A \cos C] = 4 \sin B \cos A \cos C $$

Calculamos $M$:
$$ M = \frac{4 \sin C \cos A \cos B}{4 \sin B \cos A \cos C} = \frac{\sin C \cos B}{\sin B \cos C} = \left(\frac{\sin C}{\cos C}\right) \left(\frac{\cos B}{\sin B}\right) $$
$$ M = \tan C \cot B $$

4. Resultado final:
$$ M = \tan C \cot B $$