Let \(u = \theta + \alpha\). Then \(\theta = u - \alpha\) and \(\theta + 2\alpha = u + \alpha\).

\(\sin(u - \alpha) = 3\sin(u + \alpha)\)

\(\sin u\cos\alpha - \cos u\sin\alpha = 3\sin u\cos\alpha + 3\cos u\sin\alpha\)

\(-2\sin u\cos\alpha = 4\cos u\sin\alpha \Rightarrow \tan u = -2\tan\alpha\)

So \(\tan(\theta + \alpha) + 2\tan\alpha = 0\).

Answer: (D) 0