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┬а+ c2┬атАУ a2)/2bc
2bc cos A = (b2┬а+ c2┬атАУa2) тАж (i)
Cos B = (a2┬а+ c2┬атАУ b2)/2ac
2ac cos B = (a2┬а+ c2┬атАУb2)тАж (ii)
Cos C = (a2┬а+ b2┬атАУ c2)/2ab
2ab cos C = (a2┬а+ b2┬атАУc2) тАж (iii)
Now let us add equation (i), (ii) and (ii) we get,
2bc cos A + 2ac cos B + 2ab cos C = (b2┬а+c2┬атАУ a2) + (a2┬а+ c2┬атАУb2) + (a2┬а+ b2┬атАУ c2)
Upon simplification we get,
= c2┬а+ b2┬а+ a2
2 (bc cos A + ac cos B + ab cos C) = a2┬а+b2┬а+ c2
Hence proved.