Question -
Answer -
Let us consider LHS:
2 (bc cos A + ca cos B + ab cos C)
As LHS contain 2ca cos B, 2ab cos C and 2cb cos A,which can be obtained from cosine formulae.
From cosine formula we have:
Cos A = (b2 + c2 – a2)/2bc
2bc cos A = (b2 + c2 –a2) … (i)
Cos B = (a2 + c2 – b2)/2ac
2ac cos B = (a2 + c2 –b2)… (ii)
Cos C = (a2 + b2 – c2)/2ab
2ab cos C = (a2 + b2 –c2) … (iii)
Now let us add equation (i), (ii) and (ii) we get,
2bc cos A + 2ac cos B + 2ab cos C = (b2 +c2 – a2) + (a2 + c2 –b2) + (a2 + b2 – c2)
Upon simplification we get,
= c2 + b2 + a2
2 (bc cos A + ac cos B + ab cos C) = a2 +b2 + c2
Hence proved.