If tanθ=(asinx+bsiny)/(acosx+bcosy) then show that asin(θ–x)+bsin(θ-y)=0

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if tanθ=(asinx+bsiny)/(acosx+bcosy) then show that asin(θ–x)+bsin(θ-y)=0

⇒tanθ=(asinx+bsiny)/(acosx+bcosy)

⇒sinθ/cosθ=(asinx+bsiny)/(acosx+bcosy)

⇒sinθ.acosx+sinθ.bcosy=cosθ.asinx+cosθ.bsiny

⇒sinθ.acosx+sinθ.bcosy-cosθ.asinx-cosθ.bsiny=0

⇒a(sinθ.cosx-cosθ.sinx)+b(sinθ.cosy-cosθ.siny)=0

[sin(A-B)=sinAcosB-cosA.sinB]

⇒asin(θ–x)+bsin(θ-y)=0

Hence If tanθ=(asinx+bsiny)/(acosx+bcosy) then asin(θ–x)+bsin(θ-y)=0 proved

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