If a, b, c are the sides of triangle ABC, then what is \begin{vmatrix} a^{2} & b \sin A & c \sin A \\ b \sin A & 1 & \cos A \\ c \sin A & \cos A & 1 \end{vmatrix} equal to?

Explanation

Expanding the determinant: a^2(1 - \cos^2 A) - b \sin A(b \sin A - c \sin A \cos A) + c \sin A(b \sin A \cos A - c \sin A). This simplifies to a^2 \sin^2 A - b^2 \sin^2 A - c^2 \sin^2 A + 2bc \sin^2 A \cos A = \sin^2 A (a^2 - b^2 - c^2 + 2bc \cos A). By the cosine rule, a^2 = b^2 + c^2 - 2bc \cos A, so the bracket evaluates to 0.

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