If A,B,C are the angles of a triangle , the system of equations  sinAx+y+z=cos A,x+sinBy+z=cosB, x+y+sin Cz=1−cos C has

Let Δ=sin⁡A111sin⁡B111sin⁡C Applying C2→C2−C1 and C3→C3−C1 we get Δ=sin⁡A1−sin⁡A1−sin⁡A1sin⁡B−1010sin⁡C−1 Expanding along C3, we get Δ=(1−sin⁡A)1sin⁡B−110+(sin⁡C−1)sin⁡A1−sin⁡A1sin⁡B−1=(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)Since A,B,C are the angles of a triangle  0< sinA,  sin B,  sin C≤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.