$\large{\tg(\frac{\alpha}{2}) = \sqrt{\frac{1-cos(\alpha)}{1+cos(\alpha)} \space}}$ e $\large{\tg(\frac{\beta}{2}) = \sqrt{\frac{1-cos(\beta)}{1+cos(\beta)} \space}}$ , assim: $\large{\tg(\frac{\alpha}{2})} \cdot \large{\tg(\frac{\alpha}{2})}$ $=$ $\large{\sqrt{\frac{1-cos(\alpha)}{1+cos(\alpha)} \cdot \frac{1-cos(\beta)}{1+cos(\beta)}} \space}$ . Sabendo disso, agora temos: $2\sin (\alpha) \cos (\beta) \space + \space 2\cos (\alpha) \sin (\beta) $ $=$ $\sin (\alpha) + \sin (\beta)$ Se considerarmos $\cos (\alpha) = \cos (\beta) = \large{\frac{1}{2}}$ , a igualdade permanece verdadeira sem ferir as condições propostas pelo enunciado. Desse modo: $\large{\tg(\frac{\alpha}{2})} \cdot \large{\tg(\frac{\alpha}{2})}$ $=$ $\large{\sqrt{\frac{1-\frac{1}{2}}{1+\frac{1}{2}} \cdot \frac{1-\frac{1}{2}}{1+\frac{1}{2}}} \space}$ $=$ $\large{\sqrt{\left(\frac{1}{3} \right)^2 \space}}$ $=$ $\boxed{\frac{1}{3}}$ Alternativa $\mathbb{(A)}$