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2015-05-18T12:38:36+05:30

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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

   =>  k = +1  or -1

2 5 2
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