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Question

If msinθ=n sin(θ+2α),provethattan(θ+α)cotα=m+nmn

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Solution

Given that msinθ=n sin(θ+2α)sin(θ+2α)sinθ=mn
Using componendo and dividendo, we get
sin(θ+2α)+sinθsin(θ+2α)sinθ=m+nmn2sin(θ+2α+θ2)cos(θ+2αθ2)2cos(θ+2α+θ2)sin(θ+2αθ2)=m+nmn[ sin x+siny=2sinx+y2cosxy2 and sinxsiny=2cosx+y2sinxy2]sin(θ+α)cosαcos(θ+α)sinα=m+nmntan(θ+α)cotα=m+nmn


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