If the median of ΔABC through A is perpendicular to AB then
tanA+tanB=0
2tanA+tanB=0
tanA+2tanB=0
None of these
We have BD=DC and ∠DAB=900 Draw CN perpendicular to BA produced, then in ΔBCN, we have DA=12CN and AB=AN Let ∠CAN=α∵tanA=tan(π−α)=−tanα=−CNNA=−2ADAB=−2tanB⇒tanA+2tanB=0