If xy + yz + zx = 1show that[x/(1 - x^2)] + [y/(1 - y^2)] + [z/(1 - z^2)] = (4xyz)/[(1 - x^2)(1 - y^2)(1 - z^2)]

Assuming that you know conditional identities .If not just go through them.Let x= tan A, y=tan B ,z=tan CThen xy+yz+zx=1 gives tanA tanB +tanBtanC+tanCtanA-1=0 ----(1)Now tan (A+B+C)== tanA+tanB+tanC-tanAtanBtanC/(1-tanAtanB-tanBtanC-tanCtanA)=> tan (A+B+C)= tanA+tanB+tanc-tanAtanBtanC/(0) -----from eqn (1)tan (A+B+C) =infinity = tanPi/2 ---- coz smthng upon 0 is infinity.Thus implies A+B+C =pi/2 Or 2A+2B+2C = piOr tan (2A+2B+2C)=tan pi=0Or tan2A+tan2B+tan2C=tan2Atan2Btan2COr 2x/(1-x^2) + 2y/(1-y^2) + 2z/(1-z^2)=8xyz/[(1-x^2)(1-y^2)(1-z^2) Please hit approved button.. _/\_Now take out 2 common from both sides and u get the answer...