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
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