Question

TanA+TanB+Tanc-sin(A+B+C)/cosAcosBcosC=?

Answer 1

Answer:

TanATanBTanC.

Step-by-step explanation:

Sin(A+B+C) = Sin(A+B)CosC+ Cos(A+B)SinC

             = (SinACosB+CosASInB)CosC + (CosACosB - SinASinB)SinC

              = SinACosBCosC+ CosASinBCosC + CosACosBSinC - SinASinBSinC

Divide by CosACosCosBCosC on both sides

Sin(A+B+C)/CosACosBCosC = TanA + TanB+TanC - TanATanBTanC.

=> TanA + TanB + TanC - Sin(A+B+C)/CosACosBCosC = TanATanBTanC.

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