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

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

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

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

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

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