Come si dimostra tan (A + B + C) = (tanA + tanB + tanC-tanAtanBtanC) / (1-tanAtanB-tanBtanC-tanCtanA) tan(A+B+C)=tanA+tanB+tanC−tanAtanBtanC1−tanAtanB−tanBtanC−tanCtanA?
LHS=tan(A+B+C) LHS=tan(A+B+C)
=(tan(A+B)+tanC) /(1-tan(A+B)tanC)=tan(A+B)+tanC1−tan(A+B)tanC
=((tanA+tanB)/(1-tanAtanB)+tanC) /(1-((tanA+tanB)/(1-tanAtanB))tanC)=tanA+tanB1−tanAtanB+tanC1−(tanA+tanB1−tanAtanB)tanC
=((1-tanAtanB)((tanA+tanB)/(1-tanAtanB)+tanC)) /(((1-tanAtanB)(1-((tanA+tanB)/(1-tanAtanB))tanC))=(1−tanAtanB)(tanA+tanB1−tanAtanB+tanC)((1−tanAtanB)(1−(tanA+tanB1−tanAtanB)tanC))
= (tanA+tanB+tanC-tanAtanBtanC)/(1-tanAtanB-tanBtanC-tanCtanA)=RHS=tanA+tanB+tanC−tanAtanBtanC1−tanAtanB−tanBtanC−tanCtanA=RHS