How does cos^2A+cos^2B+cos^2C+2cosAcosBcosC=1,where A,B,C are angles of triangles?

   Dear Aparna Tiwari,       

          2cosAcosBcosC =(2cosAcosB)cosC                               here A+B+C=pi=∏       thereforen  C=pi-(A+B)
                                =(cos(A + B) + cos(A - B))cosC
                                =cos(pi - C)cosC + cos(A - B)cos(pi - (A + B))
                                =-2cos^2C - 2cos(A + B)cos(A - B)

                                =-cos^2C - cos^2A + sin^2B

cos^2A+cos^2B+cos^2C+2cosAcosBcosC=cos^2A+cos^2B+cos^2C-cos^2C - cos^2A + sin^2B

                                                                              =   cos^2B+ sin^2B =1

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