A.MATHS

tanAtanB + tanBtanC +tanCtanA
=tanAtan[90-(A+C)]+tan[90-(A+C)]tanC+tanCtanA
=tanAcot(A+C)+cot(A+C)tanC+tanCtanA
=cot(A+C)(tanA+tanC)+tanAtanC
=(1-tanAtanC)(tanA+tanC)/(tanA+tanC) +tanAtanC
=1

LHS
=tanAtanB + tanBtanC + tanCtanA
=tanB (tanA+tanC) + tanCtanA
=tan (90-A-C) * (tanA+tanC) + tanCtanA
=cot (A+C) * (tanA+tanC) + tanCtanA
=cot (A+C) * [tanA+tanC] (1-tanAtanC) + tanCtanA
=1 - tanAtanC + tanCtanA
=1
=RHS

First, tan C = tan[90degree -(A+B)]= sin [90degree-(A+B)]/cos[90degree-(A+B)]=cos(A+B)/sin(A+B) = 1/tan(A+B) = [1-tanAtanB]/[tanA+tanB] -------- (1)
Therefore, LHS = tanAtanB+ tanBtanC+tanCtanA
=tanAtanB+ tanC[tanA+tanB]
= tanAtanB + {[1-tanAtanB]/[tanA+tanB]}[tanA+tanB] (because of (1)
=tanAtanB+1 -tanAtanB
=1 = RHS