If sinAsinBsinC=p and cosAcosBcosC=q, where A,B,C are the angles of a ΔABC, then tanAtanB+tanBtanC+tanCtanA is equal to
For triangle ABC, A+B+C=π
⇒−1=cos(A+B+C)
⇒−1=cos((A+B)+C)
⇒−1=cos(A+B)cosC−sin(A+B)sinC
⇒−1=(cosAcosB−sinAsinB)cosC−((sinAcosB+cosAsinB)sinC)
⇒−1=cosAcosBcosC−sinAsinBcosC−sinAcosBsinC−cosAsinBsinC
⇒−1=cosAcosBcosC(1−sinAsinBcosAcosB−sinAsinCcosAcosC−sinCsinBcosCcosB)
⇒−1=q(1−tanAtanB−tanAtanC−tanBtanC)
⇒−1q=1−(tanAtanB+tanAtanC+tanBtanC)
⇒tanAtanB+tanAtanC+tanBtanC=1+1q
⇒tanAtanB+tanAtanC+tanBtanC=q+1q