2016/052) if $n\sin \theta =\sin (\theta+ 2a)$ then $\tan (\theta+a) = ?$

$\frac{\sin (\theta+2a)}{\sin\theta} = n$
using componendo dividendo we have
$\frac{\sin (\theta+2a) +  \sin \theta}{\sin (\theta+2a) -  \sin \theta} = \frac{n+1}{n-1}$
Or $\frac{2 \sin (\theta+a)  \cos\,  a}{2 \cos (\theta+a)  \sin\, a} = \frac{n+1}{n-1}$
Or $\frac{tan  (\theta+a)}{\tan\,   a} = \frac{n+1}{n-1}$

Or  $\tan  (\theta+a) =    \frac{n+1}{n-1} \tan\, a$