If A+B+C =`pi`, show that `|["sin"^(2)A, "sin A cos A", "cos"^(2)A],["sin"^(2) B, "sin B cos B", "cos"^(2)B],["sin"^(2)C, "sin C cos C", "cos"^(2)C]|

Let the given determinant be `Delta.` Then,
`Delta = |[1, (1//2)"sin"2A, (1//2)(1+"cos"2A)],[1,(1//2)"sin"2B , (1//2)(1+"cos"2B)],[1, (1//2)"sin" 2C, (1//2)(1+"cos"2C)]|[C_(1) to C_(1) + C_(3)]`
`= (1)/(4)*|[1, "sin"2A, 1+"cos"2A],[0,"sin"2B , 1+"cos"2B],[1, "sin" 2C, 1+"cos"2C]|`
`= (1)/(4)*|[1, "sin"2A, 1+"cos"2A],[0,"sin"2B-"sin"2A, "cos"2B-"cos"2A],[0, "sin"2C-"sin"2A, "cos"2C-"cos"2A]| {[R_(2) to R_(2)-R_(1)],[R_(3) to R_(3) -R_(1)]}`
`= (1)/(4)*|[1, "sin"2A, 1+"cos"2A],[0,2"cos"(A+B)"sin"(B-A), 2"sin"(A+B)"sin"(A-B)],[0, 2"cos"(A+C)"sin"(C-A), 2"sin"(A+C)"sin"(A-C)]|`
`= "sin"(A-B)"sin"(A-C)*|[1, "sin"2A, 1+"cos"2A],[0,-"cos"(A+B),"sin"(A+B)],[0, -"cos"(A+C), "sin"(A+C)]|["taking 2sin(A-B) common from "R_(2) "and 2sin (A-C) common from"R_(3)]`
`= "sin"(A-B)"sin"(A-C)*|[1, "sin"2A, 1+"cos"2A],[0,-"cos"(pi-C),"sin"(pi-C)],[0, -"cos"(pi-B), "sin"(pi-B)]|[because A+B+C = pi]`
`= "sin"(A-B)"sin"(A-C)*|[1, "sin"2A, 1+"cos"2A],[0,"cos"C,"sin"C],[0, "cos"B, "sin"B]|`
`="sin"(A-B)"sin"(A-C)["sin"B "cos"C- "cos B sin C"]`
`= "sin (A-B) sin (A-C) sin (B-C)"`
`=-"sin(A-B) sin (B-C) sin (C-A)"`
Hence, `Delta, = -"sin (A-B) sin (B-C) sin (C-A)"`