If xtan(θ+A)=ytan(θ+B)=ztan(θ+C)  then ∑x+yx−ysin2(A−B)=

Let θ=00 .  Then                xtanA=ytanB=ztanC=k                ∑tanA+tanBtanA−tanBsin2(A−B)                =∑sin(A+B)sin(A−B).sin2(A−B)                =∑sin(A+B).sin(A−B)                =∑sin2A−sin2B                =sin2A−sin2B+sin2B−sin2C+sin2C−sin2A=0