Question

If the median of a triangle $ABC$ passing through $A$ is perpendicular to $AB,$ then $\tan A+2\tan B=$

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$


Let $D$ be the point where median through $A$ meets the side $BC$. Then we have:
$\angle DAC=\angle A-{90}^0,\angle CDA={90}^0+B,$
Now, applying sine rule in $\Delta ADC$ we have,
$\frac{\frac{a}{2}}{\sin (A-{90}^{\circ })}=\frac{b}{\sin ({90}^{\circ }+B)}$
$\Rightarrow \frac{a}{-2\cos A}=\frac{b}{\cos B}$
Writing $a$ as $2R\sin A$ and $b$ as $2R\sin B$
$\Rightarrow \frac{2R\sin A}{-2\cos A}=\frac{2R\sin B}{\cos B}$
$\Rightarrow \frac{\sin A}{-2\cos A}=\frac{\sin B}{\cos B}$
$\Rightarrow \frac{-\tan A}{2}=\tan B$
$\Rightarrow -\tan A=2\tan B\Rightarrow \tan A+2\tan B=0$