Mark the correct alternative in each of the following :
In any ΔABC, 2(bc cos A+ca cos B+ab cos C)=
a2+b2+c2
2bc cos A+2ca cos B+2ab cos C ...(i)Putting the values of cos A, cos B and cos C in (i), we get=2bc(b2+c2−a22bc)+2ca(c2+a2−b22ca)+2ab(a2+b2−c22ab)=(b2+c2−a2)+(c2+a2−b2)+(a2+b2−c2)=a2+b2+c2