Let AD,BE,CF be the lengths of internal bisectors of angles A,B,C respectively of triangle ABC Then the harmonic mean of $A D \sec \frac{A}{2}, B E \sec \frac{B}{2}, C F \sec \frac{C}{2}$ equal to

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Sum of reciprocals of the sides of triangle ABC

Harmonic mean of sides of triangle ABC

Arithmetic mean of sides of triangle ABC

Geometric mean of sides of triangle ABC

answer is B.

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Detailed Solution

Length of angle bisector $A D = \frac{2 b c}{b + c} \cos \frac{A}{2}$
Length of angle bisector $B E = \frac{2 a c}{a + c} \cos \frac{B}{2}$
Length of angle bisector $C F = \frac{2 a b}{a + b} \cos \frac{C}{2}$
$H.M. = \frac{3}{\frac{b + c}{2 b c} + \frac{a + c}{2 a c} + \frac{a + b}{2 a b}} = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$