If `cos(alpha-beta)+cos(beta-gamma)+cos(gamma-alpha)=-3/2`, prove that`cosalpha+cosbeta+cosgamma=sinalpha+sinbeta+singamma=0`

If `cos(alpha-beta)+cos(beta-gamma)+cos(gamma-alpha)=-3/2`, prove that

1.	If `cos(alpha-beta)+cos(beta-gamma)+cos(gamma-alpha)=-3/2`, prove that`cosalpha+cosbeta+cosgamma=sinalpha+sinbeta+singamma=0`
Answer» `cos(alpha-beta)+cos(beta-gamma)+cos(gamma-alpha) = -3/2` `=>2(cosalphacosbeta+sinalphasinbeta)+2(cosbetacosgamma+sinbetasingamma)+2(cosalphacosgamma+sinalphasingamma) = -3` `=>2(cosalphacosbeta+cosbetacosgamma+cosalphacosgamma)+2(sinalphasinbeta+sinbetasingamma+sinalphasingamma)+3 = 0` `=>2(cosalphacosbeta+cosbetacosgamma+cosalphacosgamma)+2(sinalphasinbeta+sinbetasingamma+sinalphasingamma)+(cos^2alpha+sin^2alpha)+(cos^2beta+sin^2beta)+(cos^2gamma+sin^2gamma) = 0` `=>(cos^2alpha+cos^2beta+cos^2gamma +2(cosalphacosbeta+cosbetacosgamma+cosalphacosgamma))+(sin^2alpha+sin^2beta+sin^2gamma +2(sinalphasinbeta+sinbetasingamma+sinalphasingamma))=0` `=>(cosalpha+cosbeta+cosgamma)^2+(sinalpha+sinbeta+singamma)^2 = 0` `a^2+b^2 = 0` is possible onle when `a = b = 0.` `:. cosalpha+cosbeta+cosgamma = sinalpha+sinbeta+singamma= 0`