In a triangle $A B C$ , the median $A D$ is perpendicular to the side $A B$ , then

2 tan A + tan B = 0

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tan A + 2 tan B = 0

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tan A - 2 tan B = 0

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2 tan A - tan B = 0

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Solution

The correct option is D $tan A + 2 tan B = 0$

$θ = B + 90 °$
$(1 + 1) cot (B + 90 °) = cot 90 ° - cot (A - 90 °)$
$- 2 tan B = 0 + tan A$
$tan A + 2 tan B = 0$
Hence, $tan A + 2 tan B = 0$ is the correct answer.

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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

∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 cot \frac{A}{2} & cot \frac{B}{2} & cot \frac{C}{2} tan \frac{B}{2} + tan \frac{C}{2} & tan \frac{C}{2} + tan \frac{A}{2} & tan \frac{A}{2} + tan \frac{B}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0

△ A B C

∣ ∣ ∣ ∣ ∣ \begin{matrix} cot \frac{A}{2} & cot \frac{B}{2} & cot \frac{C}{2} tan \frac{B}{2} + tan \frac{C}{2} & tan \frac{A}{2} + tan \frac{C}{2} & tan \frac{A}{2} + tan \frac{B}{2} 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

tan \frac{A}{2} tan \frac{B}{2} + tan \frac{B}{2} tan \frac{C}{2} + tan \frac{C}{2} tan \frac{A}{2} = 1

In Δ ABC, it is being given that

Δ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 cot \frac{A}{2} & cot \frac{B}{2} & cot \frac{C}{2} tan \frac{B}{2} + tan \frac{C}{2} & tan \frac{A}{2} + tan \frac{C}{2} & tan \frac{A}{2} + tan \frac{B}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 0,

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