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In any $\triangle A B C, \frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=$
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Verified Answer
The correct answer is:
$\frac{a^2+b^2+c^2}{2 a b c}$
Given, in a $\triangle A B C$,
$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$
We know that,
$\begin{aligned} & \cos A=\frac{b^2+c^2-a^2}{2 b c}, \\ & \cos B=\frac{a^2+c^2-b^2}{2 a c} \\ & \cos C=\frac{a^2+b^2-c^2}{2 a b}\end{aligned}$
Since, $\frac{b^2+c^2-a^2}{2 a b c}+\frac{a^2+c^2-b^2}{2 a b c}+\frac{a^2+b^2-c^2}{2 a b}$
$=\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2 a b c}$
$=\frac{a^2+b^2+c^2}{2 a b c}$
$\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}$
We know that,
$\begin{aligned} & \cos A=\frac{b^2+c^2-a^2}{2 b c}, \\ & \cos B=\frac{a^2+c^2-b^2}{2 a c} \\ & \cos C=\frac{a^2+b^2-c^2}{2 a b}\end{aligned}$
Since, $\frac{b^2+c^2-a^2}{2 a b c}+\frac{a^2+c^2-b^2}{2 a b c}+\frac{a^2+b^2-c^2}{2 a b}$
$=\frac{b^2+c^2-a^2+a^2+c^2-b^2+a^2+b^2-c^2}{2 a b c}$
$=\frac{a^2+b^2+c^2}{2 a b c}$
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