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Question
With usual notations prove that 2(bc cosA + ac cosB + ab cosC) = a2 + b2 + c2 .
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Solution
L.H.S. = 2(bc cosA + ac cosB + ab cosC)
= 2bc cosA + 2ac cosB + 2ab cosC
`= 2bc((b^2 + c^2 - a^2)/(2bc)) + 2ac((c^2 + a^2 - b^2)/(2ca)) + 2ab((a^2 + b^2 - c^2)/(2ab))` ...[By cosine rule]
= b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2
= a2 + b2 + c2
= R.H.S.
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