If A, B, C are the angles of a triangle,then | vedclass

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(B) Let the determinant be $\Delta = \left| \begin{array}{ccc} -1 & \cos C & \cos B \ \cos C & -1 & \cos A \ \cos B & \cos A & -1 \end{array} ight

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(B) Let the determinant be $\Delta = \left| \begin{array}{ccc} -1 & \cos C & \cos B \ \cos C & -1 & \cos A \ \cos B & \cos A & -1 \end{array} ight|$.Expanding along the first row:$\Delta = -1(1 - \cos^2 A) - \cos C(-\cos C - \cos A \cos B) + \cos B(\cos C \cos A + \cos B)$$\Delta = -\sin^2 A + \cos^2 C + \cos A \cos B \cos C + \cos A \cos B \cos C + \cos^2 B$$\Delta = -\sin^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C$.Since $A+B+C = \pi$,we have $A = \pi - (B+C)$,so $\cos A = -\cos(B+C) = -(\cos B \cos C - \sin B \sin C) = \sin B \sin C - \cos B \cos C$.Substituting $\cos A$ into the expression:$\Delta = -\sin^2 A + \cos^2 B + \cos^2 C + 2 \cos B \cos C (\sin B \sin C - \cos B \cos C)$$\Delta = -\sin^2 A + \cos^2 B + \cos^2 C + 2 \cos B \cos C \sin B \sin C - 2 \cos^2 B \cos^2 C$Using $\sin^2 A = \sin^2(B+C) = (\sin B \cos C + \cos B \sin C)^2 = \sin^2 B \cos^2 C + \cos^2 B \sin^2 C + 2 \sin B \cos C \cos B \sin C$.Substituting this back,all terms cancel out to give $\Delta = 0$.

If $A, B, C$ are the angles of a triangle,then $\left| \begin{array}{ccc} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{array} \right| = $

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