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Question

If tanβ=tanα+tanγ1+tanαtanγ, then prove that sin2β=sin2α+sin2γ1+sin2αsin2γ.

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Solution

tanβ=tanα+tany1+tanαtany


=sinαcosα+sinycosy1+sinαcosαsinycosy


=sin(α+y)cos(αy)


Now sin2β=2tanβ1+tan2β


=2⎢ ⎢ ⎢ ⎢ ⎢sin(α+y)cos(αy)1+sin2(α+y)cos2(αy)⎥ ⎥ ⎥ ⎥ ⎥


=2[sin(α+y)cos(αy)(cosαcosy+sinαsiny)2+(sinαcosy+cosαsiny)2]


=2⎢ ⎢ ⎢ ⎢ ⎢(sinαcosy+cosαsiny)(cosαcosy+sinαsiny)cos2αcos2y+sin2αsin2y+2sinαcosαcosysiny+sin2αcos2y+cos2αsin2y+2cosαcosysinαsiny⎥ ⎥ ⎥ ⎥ ⎥


=2[sinαcosαcos2y+sin2αcosysiny+cos2αsinycosy+sin2ycosαsinαcos2y+sin2y+4sinαcosαcosysiny]


=2[sinαcosα+sinycosy1+sin2αsin2y]


=sin2α+sin2y1+sin2αsin2y=RHS.

Hence LHS=RHS proof


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