If $α, β, γ$ are lengths of internal bisectors of angles A, B, C respectively of $∆ ABC$ , then $\sum \frac{1}{α} \cos \frac{A}{2}$ is

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a + b + c

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

$2 (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

2(a+b+c)

answer is B.

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Detailed Solution

$\begin{array}{l} α = \frac{2 b c}{b + c} \cos \frac{A}{2} \\ β = \frac{2 a c}{a + c} \cot \frac{β}{2} \\ γ = \frac{2 a b}{a + b} \cos \frac{c}{2} \\ \sum \frac{1}{α} \cos \frac{A}{2} = \frac{1}{α} \cos \frac{A}{2} + \frac{1}{β} \cos \frac{β}{2} + \frac{1}{γ} \cos \frac{c}{2} \begin{array}{l} = \frac{b + c}{2 b c} + \frac{a + c}{2 a c} + \frac{a + b}{2 a b} \\ = \frac{1}{2} [\frac{b}{b c} + \frac{c}{b c}] + \frac{1}{2} [\frac{a}{a c} + \frac{c}{a c}] + \frac{1}{2} [\frac{a}{a b} + \frac{b}{a b}] \\ = \frac{1}{2} [(\frac{1}{c} + \frac{1}{b} + \frac{1}{c} + \frac{1}{a} + \frac{1}{b} + \frac{1}{a})] \\ = \frac{1}{2} [\frac{2}{c} + \frac{2}{b} + \frac{2}{a}] \\ = \frac{2}{2} [\frac{1}{a} + \frac{1}{b} + \frac{1}{c}] \\ = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \end{array} \end{array}$