If a,b,cϵR and a2+b2−ab−a−b+1≤0 and α+β+γ=0, then Δ=∣∣
∣
∣∣1cosγcosβcosγacosαcosαcosβb∣∣
∣
∣∣ equals
A
ab
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B
1
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C
0
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D
-1
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Solution
The correct option is A 0 a2+b2−ab−a−b+1≤0⇒a2+b2−2ab+ab−a−b+1≤0⇒a=b=1 As α+β+γ=0 Let α=B−C,β=C−A,γ=A−B Δ=∣∣
∣
∣∣1cos(A−B)cos(C−A)cos(A−B)1cos(B−C)cos(B−C)cos(C−A)1∣∣
∣
∣∣