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if tan(alpha+theta) = ntan (alpha-theta) then show that : (n+1)sin 2theta=(n-1)sin 2alpha

$\tan (α + θ) = n \tan (α - θ) \Rightarrow \frac{\tan α + \tan θ}{1 - \tan α \tan θ} = n (\frac{\tan α - \tan θ}{1 + \tan α \tan θ}) \Rightarrow (\tan α + \tan θ) (1 + \tan α \tan θ) = n (\tan α - \tan θ) (1 - \tan α \tan θ) \Rightarrow \tan α + \tan θ + \tan^{2} α \tan θ + \tan α \tan^{2} θ = n \tan α - n \tan θ - n \tan^{2} α \tan θ + n \tan α \tan^{2} θ R e a r r a n g i n g t h e t e r m s, w e h a v e \Rightarrow \tan α + \tan α \tan^{2} θ + \tan θ + \tan^{2} α \tan θ = n \tan α + n \tan α \tan^{2} θ - n \tan θ - n \tan^{2} α \tan θ \Rightarrow \tan α (1 + \tan^{2} θ) + \tan θ (1 + \tan^{2} α) = n \tan α (1 + \tan^{2} θ) - n \tan θ (1 + \tan^{2} α) \Rightarrow \tan θ (1 + \tan^{2} α) + n \tan θ (1 + \tan^{2} α) = n \tan α (1 + \tan^{2} θ) - \tan α (1 + \tan^{2} θ) \Rightarrow (n + 1) \tan θ (1 + \tan^{2} α) = (n - 1) \tan α (1 + \tan^{2} θ) \Rightarrow (n + 1) \frac{\tan θ}{1 + \tan^{2} θ} = (n - 1) \frac{\tan α}{1 + \tan^{2} α} \Rightarrow (n + 1) \frac{2 \tan θ}{1 + \tan^{2} θ} = (n - 1) \frac{2 \tan α}{1 + \tan^{2} α} \Rightarrow (n + 1) \sin 2 θ = (n - 1) \sin 2 α H e n c e p r o v e d .$

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