If a, b, c are the sides of a ∆ ABC opposite angles A,B,C respectively, and Δ=a2bsin⁡Acsin⁡Absin⁡A1cos⁡(B−C)csin⁡Acos⁡(B−C)1, then ∆equals

By the law of sinesasin⁡A=bsin⁡B=csin⁡C=k(say)⇒a=ksin⁡A,b=ksin⁡B,c=ksin⁡C. Now Δ=a2ab/kac/kab/k1cos⁡(B−C)ac/kcos⁡(B−C)1=a21sin⁡Bsin⁡Csin⁡B1cos⁡(B−C)sin⁡Ccos⁡(B−C)1=a21sin⁡(A+C)sin⁡(A+B)sin⁡(A+C)1cos⁡(B−C)sin⁡(A+B)cos⁡(B−C)1=a2sin⁡Acos⁡A0cos⁡Csin⁡C0cos⁡Bsin⁡B0sin⁡Acos⁡A0cos⁡Csin⁡C0cos⁡Bsin⁡B0=a2(0)=0