1. Cálculo de $m^2 - n^2$:
Usamos diferencia de cuadrados: $m^2 - n^2 = (m+n)(m-n)$.
$$ m+n = (\tan \theta + \sin \theta) + (\tan \theta - \sin \theta) = 2\tan \theta $$
$$ m-n = (\tan \theta + \sin \theta) - (\tan \theta - \sin \theta) = 2\sin \theta $$
Entonces: $m^2 - n^2 = (2\tan \theta)(2\sin \theta) = 4 \tan \theta \sin \theta$.

2. Cálculo de $4\sqrt{mn}$:
$$ mn = (\tan \theta + \sin \theta)(\tan \theta - \sin \theta) = \tan^2 \theta - \sin^2 \theta $$
$$ mn = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta \left( \frac{1}{\cos^2 \theta} - 1 \right) = \sin^2 \theta \left( \frac{1 - \cos^2 \theta}{\cos^2 \theta} \right) $$
$$ mn = \sin^2 \theta \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right) = \sin^2 \theta \tan^2 \theta $$
Tomando la raíz cuadrada:
$$ 4\sqrt{mn} = 4\sqrt{\sin^2 \theta \tan^2 \theta} = 4 \sin \theta \tan \theta $$

3. Conclusión:
Ambos lados son iguales a $4 \tan \theta \sin \theta$.
$$ \boxed{m^2 - n^2 = 4\sqrt{mn}} $$