In Delta ABC,frac{cos A}{a} + frac{cos B}{b} | vedclass

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Question

(B) Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.Sub

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$,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 $\Delta ABC$,$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = $

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