Come si dimostra tan (A + B + C) = (tanA + tanB + tanC-tanAtanBtanC) / (1-tanAtanB-tanBtanC-tanCtanA) tan(A+B+C)=tanA+tanB+tanCtanAtanBtanC1tanAtanBtanBtanCtanCtanA?

LHS=tan(A+B+C) LHS=tan(A+B+C)

=(tan(A+B)+tanC) /(1-tan(A+B)tanC)=tan(A+B)+tanC1tan(A+B)tanC

=((tanA+tanB)/(1-tanAtanB)+tanC) /(1-((tanA+tanB)/(1-tanAtanB))tanC)=tanA+tanB1tanAtanB+tanC1(tanA+tanB1tanAtanB)tanC

=((1-tanAtanB)((tanA+tanB)/(1-tanAtanB)+tanC)) /(((1-tanAtanB)(1-((tanA+tanB)/(1-tanAtanB))tanC))=(1tanAtanB)(tanA+tanB1tanAtanB+tanC)((1tanAtanB)(1(tanA+tanB1tanAtanB)tanC))

= (tanA+tanB+tanC-tanAtanBtanC)/(1-tanAtanB-tanBtanC-tanCtanA)=RHS=tanA+tanB+tanCtanAtanBtanC1tanAtanBtanBtanCtanCtanA=RHS

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