If $\sin\theta=n\sin(\theta+2\alpha)$ then $tan(\theta+\alpha)$ is equal to

Question

$\begin{array}{1 1}(a)\;\large\frac{1+n}{2-n}\normalsize\tan\alpha&(b)\large\frac{1-n}{1+n}\normalsize\tan\alpha\\(c)\;\tan\alpha&(d)\;\large\frac{1+n}{1-n}\normalsize\tan\alpha\end{array}$