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Sin (A + c) / Cos (B+C) = Sin (B+c) / Cos (C+A) = Sin (C + c) / Cos (A+B) = k

Sin A Cos c + Cos A Sin c = k Cos (B+C) --- (1) Sin B Cos c + Cos B Sin c = k Cos (A+ C) --- (2) Sin C Cos c + Cos C Sin c = k Cos (A + B) --- (3)

Let us find the value of Cos c and Sin c from (1) and (2).

(1) * Sin B - (2) * Sin A => Sin c Sin (B - A) = k [ Sin B Cos (B + C) - Sin A Cos (A + C) ] => = k/2 * [ -Sin C + Sin (2B+C) + Sin C - Sin (2A+C) ] = k/2 * [ - Sin (2A+C) + Sin (2B +C) ] = k * Sin (B - A) Cos(A+B+C)

=> Sin c = k Cos(A+B+C) , if A ≠ B or A-B ≠ 2π n --- (4)

Now from (1), Sin A Cos c = k [ Cos (B+C) - Cos A Cos (A+B+C) ] = k [ Cos (B+C) - Cos²A Cos (B+C) + Cos A Sin A Sin(B+C) ] = k [ Sin² A Cos(B+C) + Sin A Cos A Sin (B+C)

=> Cos c = k [ Sin A Cos (B+C) + Cos A Sin (B+C) ] , if A ≠ 2 π n or 0 => Cos c = k Sin (A+B+C) --- (5)

we know that Sin² c + Cos² c = 1 => 1 = k² [ Cos² (A+B+C) + Sin²(A+B+C) ] = k²