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Question

If tan β=tanα+tanγ1+tanαtanγ, Prove that: sin2β=sin2α+sin2γ1+sin2αsin2γ

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Solution

tanβ=sin(α+γ)cos(αγ)
on changing to sin and cos
sin2β=2tanβ1+tan2β=2sin(α+γ)cos(αγ)cos2(αγ)+sin2(α+γ)
=sin2α+sin2γsin2(α+γ)+1sin2(αγ)
=sin2α+sin2γ1+sin2αsin2γ by sin2Asin2B
using formula of sin2Asin2B

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