If the mediam of a triangle ABC through Ais perpendicular to AB, prove that tan A + TAN B =0????

dear student,

we can prove  tanA + 2tanB = 0.
Let AD be the median through A
=> AB = BDcosB
Applying sine rule to ΔABC,
AB/sinC = BC/sinA
=> BDcosB/sinC = 2BD/sinA
=> 2sinC - sinAcosB = 0
=> 2sin(A+B) - sinAcosB = 0 ... [because C = π - (A+B)]
=> 2sinAcosB + 2cosAsinB - sinAcosB = 0
=> sinAcosB + 2cosAsinB = 0
Dividing by cosAcosB,
=> tanA + 2tanB = 0.

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