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Q.

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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a

Sum of reciprocals of the sides of triangle ABC

b

Harmonic mean of sides of triangle ABC

c

Arithmetic mean of sides of triangle ABC

d

Geometric mean of sides of triangle ABC

answer is B.

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Detailed Solution

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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 $C F = \frac{2 a b}{a + b} \cos \frac{C}{2}$
$H.M. = \frac{3}{\frac{b + c}{2 b c} + \frac{a + c}{2 a c} + \frac{a + b}{2 a b}} = \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

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