If A,B,C are the angles of a triangle , the system of equations $(\sin A) x + y + z = \cos A, x + (\sin B) y + z = \cos B, x + y + (\sin C) z = 1 - \cos C$ has

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no solution

unique solution

infinitely many solutions

finitely many solutions

answer is B.

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Detailed Solution

$Let Δ = |\begin{array}{ccc} \sin A & 1 & 1 \\ 1 & \sin B & 1 \\ 1 & 1 & \sin C \end{array}|$

$Applying C_{2} \to C_{2} - C_{1} and C_{3} \to C_{3} - C_{1} we get Δ = |\begin{array}{ccc} \sin A & 1 - \sin A & 1 - \sin A \\ 1 & \sin B - 1 & 0 \\ 1 & 0 & \sin C - 1 \end{array}|$

$Expanding along C_{3}, we get Δ = (1 - \sin A) |\begin{array}{cc} 1 & \sin B - 1 \\ 1 & 0 \end{array}| + (\sin C - 1) |\begin{array}{cc} \sin A & 1 - \sin A \\ 1 & \sin B - 1 \end{array}|$

$\begin{aligned} = (1 - \sin A) (1 - \sin B) + (\sin C - 1) [\sin A (\sin B - 1) - (1 - \sin A)] \\ = \sin A (1 - \sin B) (1 - \sin C) + (1 - \sin A) (1 - \sin B) + (1 - \sin A) (1 - \sin C) \end{aligned}$

Since A,B,C are the angles of a triangle $0 < \sin A, \sin B, \sin C \leq 1$ . Also , at most one of sin A, sin B, sin C can be equal to 1.

Thus , at most two of three terms in $Δ$ can be zero and remaining must be positive . Therefore, $Δ > 0$
Hence , the given system of equations has a unique solution.

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