If a,b and c are the sides of a triangle and A, B and C are the angles opposite to a, b and c respectively, thenΔ=a2bsin⁡ACsin⁡Absin⁡A1cos⁡ACsin⁡Acos⁡A1 is independent of

∵ Δ=a2bsin⁡Acsin⁡Absin⁡A1cos⁡Acsin⁡Acos⁡A1Taking common a from each and R1 then C1, thenΔ=1bsin⁡Aacsin⁡Aabsin⁡Aa1cos⁡Acsin⁡Aacos⁡A1=1sin⁡Bsin⁡Csin⁡B1cos⁡Asin⁡Ccos⁡A1 [ by sine rule]  Applying C2→C2−sin⁡BC1 and C3→C3−sin⁡CC1, then∆=1⋯0⋯0⋮    sin B 1-sin2B cosA-sinB sinC⋮    sin C cosA-sinB sinC 1-sin2CExpanding along R1, then     Δ=cos2⁡Bcos⁡[π−(B+C)] −sin⁡Bsin⁡Ccos⁡[π−(B+C)]−sin⁡Bsin⁡Ccos2⁡C                                           (∴ A+B+C=π)      =cos2⁡B−cos⁡(B+C)−sin⁡Bsin⁡C−cos⁡(B+C)−sin⁡Bsin⁡Ccos2⁡C      =cos2⁡B−cos⁡Bcos⁡C−cos⁡Bcos⁡Ccos2⁡C=cos2⁡Bcos2⁡C−cos2⁡Bcos2⁡C=0