Prove: sin(A−B)sinAsinB+sin(B−C)sinBsinC+sin(C−A)sinCsinA=0
Given: sin(A−B)sinAsinB+sin(B−C)sinBsinC+sin(C−A)sinCsinA=0
LHS: sin(A−B)sinAsinB+sin(B−C)sinBsinC+sin(C−A)sinCsinA
We know that sin(α+β)=sinαcosβ−cosαsinβ
=sinAcosB−cosAsinBsinAsinB+sinBcosC−cosBsinCsinBsinC+sinCcosA−cosCsinAsinCsinA
=sinAcosBsinAsinB−cosAsinBsinAsinB+sinBcosCsinBsinC−cosBsinCsinBsinC+sinCcosAsinCsinA−cosCsinAsinCsinA
=cotB−cotA+cosC−cotB+cotA−cotC
=0
= RHS
Hence proved.