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sinx+siny=a and cosx+cosy=b. Prove that tan2(xy2)=4a2b2a2+b2
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Solution:

Given, a=sinx+sinyb=cosx+cosy

a2+b2=[sinx+siny]2+[cosx+cosy]2 Formula S+S=2SCC+C=2CC=[2sin(x+y2)cos(xy2)]2+2cos(x+y2)cos(xy2)]2=4sin2(x+y2)cos2(xy2)+4cos2(x+y2)cos2(xy2)4cos2(xy2) is common to both areterms=4cos2(xy2)[sin2(x+y2)+cos2(x+y2)]=4cos2(xy2)×1sin2θ+cos2θ=1a2+b2=4cos2(xy2)

 RHS =4a2b2(a2+b2)=4(a2+b2)(a2+b2)=44cos2(xy2)4cos2(xy2)=4[1cos2(xy2)]4cos2(xy2)=sin2(xy2)cos2(xy2)1cos2θ=sin2θ=tan2(xy2)= RHS 

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