Given,
xtan(θ+α)=ytan(θ+β)=ztan(θ+γ)
⇒(x+y)(x−y)=(tan(θ+α)+tan(θ+β))(tan(θ+α)−tan(θ+β))
We know that, (TanA+TanB)(TanA−TanB)=Sin(A+B)Sin(A−B)
⇒(x+y)(x−y)=sin((θ+α)+(θ+β))sin((θ+α)−(θ+β))
⇒(x+y)(x−y)=sin(2θ+α+β)sin(α−β)
Consider,sin2(α−β)×(x+y)(x−y)=sin2(α−β)×sin(2θ+α+β)sin(α−β)
=sin(α−β)×sin(2θ+α+β)
=12∗(cos(β+θ)−cos(α+θ)
Consider,sin2(α−β)∗(x+y)(x−y)+sin2(β−γ)×(y+z)(y−z)+sin2(γ−α)∗(z+x)(z−x)=12∗((cos(β+θ)−cos(α+θ))+(cos(γ+θ)−cos(β+θ))+(cos(α+θ)−cos(β+θ)))=0
Ans:1