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Antiderivative

In a Δ ABC,...

Question

In a $Δ A B C$ , if median $A D$ is perpendicular to $A B$ , then $tan A + 2 tan B$ is equal to

1

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3

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0

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\frac{1}{2}

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Solution

The correct option is C $0$
R.E.F.Image.
In $△ A C D$
By sin rule
$\frac{a}{2} \frac{s i n B}{s i n (180 - A - B)} = \frac{a}{2} / s i n (A - 90) .$
$\frac{s i n B}{s i n (A + B)} = \frac{- 1}{c o s A} .$
$S i n B c o s A = - s i n (A + B)$
$s i n B c o s A = - [s i n A c o s B + c o s A s i n B]$
$t a n A + 2 t a n B = 0.$

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Similar questions

If the median of $Δ A B C$ through $A$ is perpendicular to $A B$ then

△ A B C

t a n A + 2 t a n B = 0

Δ A B C, \frac{r}{r_{1}} = \frac{1}{2},

t a n \frac{A}{2} (t a n \frac{B}{2} + t a n \frac{C}{2})

A B C

A

A B,

tan A + 2 tan B =

Δ A B C,

A D

∠ B A C

∠ B A D : ∠ C A D = 2 : 1

(\frac{A}{3})

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