question_answer

If the median of \[\Delta ABC\]through A is perpendicular to \[AB\], then

A) \[\tan A+\tan B=0\]

B) \[2\tan A+\tan B=0\]

C) \[\tan A+2\tan B=0\]

D) None of these

Correct Answer: C

Solution :

We have \[BD=DC\]and \[\angle DAB={{90}^{o}}\]. Draw CN perpendicular to BA produced, then in \[\Delta BCN\], we have \[DA=\frac{1}{2}CN\] and \[AB=AN\] Let \[\angle CAN=\alpha \] \[\because \tan A=\tan (\pi -\alpha )=-\tan \alpha \]\[=-\frac{CN}{NA}=-2\frac{AD}{AB}=-2\tan B\] Þ \[\tan A+2\tan B=0\].