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In ∆ABC , prove that: cosA/(bcosC + ccosB) + cosB/(ccosA + acosC) + cosC/(acosB + bcosA) = (a^2 + b^2 + c^2)/2abc.

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In ∆ABC , prove that:

cosA/(bcosC + ccosB) + cosB/(ccosA + acosC) + cosC/(acosB + bcosA) = (a2 + b2 + c2)/2abc.

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From projection formula

a cosB + b cos A = C

cosA = (b2 + c2 - a2)/2bc

So L.H.S. = (b2 + c2 - a2)/abc + (c2 + a2 - b2)/abc + (a2 + b2 - c2)/abc

= (a2 + b2 + c2)/abc

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