Let $\frac { \tan ( \alpha - \beta + \gamma ) } { \tan ( \alpha + \beta - \gamma ) } = \frac { \tan \beta } { \tan \gamma }$. Then\\
(A) $\sin ( \beta - \gamma ) = \sin ( \alpha - \beta )$.\\
(B) $\sin ( \alpha - \gamma ) = \sin ( \beta - \gamma )$.\\
(C) $\sin ( \beta - \gamma ) = 0$.\\
(D) $\sin 2 \alpha + \sin 2 \beta + \sin 2 \gamma = 0$