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If sinθ = n sin(θ + 2α), then prove that tan(θ + α) =(1+n)/(1-n)tanα

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If sinθ = n sin(θ + 2α), then prove that tan(θ + α) =(1+n)/(1-n)tanα

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sinθ = n sin(θ + 2α)

=> 1/n = {sin(θ+2)}/sinθ, by componendo and dividendo we get

=

=

=

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