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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 = BDcosBApplying 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 = 0Dividing by cosAcosB,=> tanA + 2tanB = 0.
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