In any triangle ABC,frac{cos A}{a} + frac{cos | vedclass

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(B) In any $	riangle ABC$,by the Law of Cosines,we have: $\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^

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$\frac{a^2+b^2+c^2}{2abc}$

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(B) In any $	riangle ABC$,by the Law of Cosines,we have: $\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$. Substituting these into the expression: $\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc}$ $= \frac{(b^2+c^2-a^2) + (a^2+c^2-b^2) + (a^2+b^2-c^2)}{2abc}$ $= \frac{a^2+b^2+c^2}{2abc}$.

In any $\triangle ABC$,$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} =$

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