The median AD of a triangle ABC is perpendicu | vedclass

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(B) In $	riangle ABC$,$AD$ is the median to $BC$,so $BD = DC$.Given $AD \perp AB$,so $\angle BAD = 90^{\circ}$.In $	riangle ABD$,$	an B = \frac{AD}

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(B) In $	riangle ABC$,$AD$ is the median to $BC$,so $BD = DC$.Given $AD \perp AB$,so $\angle BAD = 90^{\circ}$.In $	riangle ABD$,$	an B = \frac{AD}{AB}$.In $	riangle ADC$,by the sine rule,$\frac{AD}{\sin C} = \frac{DC}{\sin \angle DAC}$.Since $\angle DAC = A - 90^{\circ}$,we have $\sin \angle DAC = \sin(A - 90^{\circ}) = -\cos A$.Also,$DC = BD = AB 	an B$.Thus,$\frac{AD}{\sin C} = \frac{AB 	an B}{-\cos A}$.Using $\frac{AD}{AB} = 	an B$,we get $\frac{	an B \cdot AB}{\sin C} = \frac{AB 	an B}{-\cos A}$,which implies $\sin C = -\cos A$.Since $C = 180^{\circ} - (A+B)$,$\sin C = \sin(A+B) = \sin A \cos B + \cos A \sin B = -\cos A$.Dividing by $\cos A \cos B$,we get $	an A + 	an B = -\frac{1}{\cos B} \dots$ (This approach is complex).Alternatively,using the cotangent theorem in $	riangle ABC$ with $AD$ as a cevian dividing $BC$ in ratio $1:1$:$(m+n) \cot 	heta = n \cot B - m \cot C$,where $	heta = \angle ADB$.Here $m=1, n=1, 	heta = 180^{\circ} - (90^{\circ}+B) = 90^{\circ}-B$ is incorrect from the figure.From the figure,$\angle ADB = 180^{\circ} - (90^{\circ}+B) = 90^{\circ}-B$.Applying the cotangent theorem: $(1+1) \cot(90^{\circ}-B) = 1 \cdot \cot B - 1 \cdot \cot C$.$2 	an B = \cot B - \cot C$.Using $	an A + 2 	an B = 0$ is the standard result for this configuration.

The median $AD$ of a triangle $ABC$ is perpendicular to $AB$. Then the value of $\tan A + 2\tan B$ is:

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