If α,β,γ are three real numbers such that α+β+γ=0, then Δ=1cos⁡γcos⁡βcos⁡γ1cos⁡αcos⁡βcos⁡α1 equals

Let  A,B and C be three real numbers such that α=B−C,β=C−A and γ=A−B, clearly α+β+γ=0. We have =cos2⁡A+sin2⁡Acos⁡Acos⁡B+sin⁡Asin⁡Bcos⁡Acos⁡B+sin⁡Asin⁡Bcos2⁡B+sin2⁡Bcos⁡Ccos⁡A+sin⁡Csin⁡Acos⁡Bcos⁡C+sin⁡Bsin⁡C    cos⁡Ccos⁡A+sin⁡Asin⁡Ccos⁡Bcos⁡C+sin⁡Bsin⁡Ccos2⁡C+sin2⁡C               =cos⁡A    sin⁡A    0cos⁡B    sin⁡B    0cos⁡C    sin⁡C    0cos⁡Asin⁡A0cos⁡Bsin⁡B0cos⁡Csin⁡C0=(0)(0)=0.