2014/101) If $\sin\theta = k \sin(\theta + 2\alpha)$ prove that $\tan (\theta + \alpha) = \dfrac{1+k}{1-k}\tan \alpha$

we have
$k = \dfrac{sin\theta}{\sin(\theta + 2\alpha)}$
using componendo dividendo
 $\dfrac{k+1}{k-1} = \dfrac{\sin\theta+\sin(\theta + 2\alpha)}{\sin\theta-\sin(\theta + 2\alpha)}$
=$\dfrac{ 2 \sin(\theta + \alpha ) \cos \theta}  {-2 \cos (\theta + \alpha) \sin \theta}$
= $\dfrac{ - \tan(\theta + \alpha )}{\tan \theta}$

hence $\dfrac{1 + k} {1- k} = \dfrac{\tan(\theta + \alpha )}{\tan \theta}$

or  $\dfrac{1 + k} {1- k}\tan \theta = \tan(\theta + \alpha) $
proved