Question

In a ∆ABC Prove that [tex] \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac} = \frac{1}{2rR} ​

Answer 1

$\huge{\underline{\sf{\red{Answer}}}}$

$\dfrac{1}{ab} + \dfrac{1}{bc} + \dfrac{1}{ac} = \dfrac{1}{2rR} \\ \\ \implies \dfrac{c + a +b}{abc} \\ \implies \dfrac{2s}{4 \triangle R } \\ \implies \dfrac{1}{2 \dfrac{\triangle}{s} R} \\ \implies \dfrac{ 1}{2rR}$

Answer 2

Given,

a,b and c are in AP

Therefore,

from theory of AP,

b +c = 2a

Now,

we have

$\frac{1}{ab} \: \: and\: \: \frac{1}{bc} \: \: and \: \: \: \frac{1}{ca}$

adding the first and last terms,

we get,

$\frac{1}{ab} + \frac{1}{ac}$

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

$= \frac{1}{a} \times \frac{b + c}{bc}$

$= \frac{b + c}{abc}$

but,

from equation

b + c = 2a

Therefore,

putting the values,

we get,

$= \frac{2a}{abc}$

$= \frac{2}{bc}$

thus we get

$\frac{1}{ab }+ \frac{1}{ca} = \frac{2}{bc}$

Hence,

$\frac{1}{ab} \: \: and \: \: \frac{1}{bc} \: \: and \: \: \frac{1}{ca} \: \: are \: in \: ap$

THUS PROVED !!!

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