With usual notations, in a triangle ABC, a cos(B - C)+b cos(C-A)+ c cos(A-B) is equal to

With usual notations, in triangle ABC, a cos (B-C)+b cos (C-A)+c cos(A-B) is equal to

With usual notations,in triangle ABC,a cos(B-C)+b cos(C-A)+c cos(A-B) is equal to (a)(abc)/(R^(2)) (b) (abc)/(4R^(2)) (c) (4abc)/(R^(2)) (d) (abc)/(2R^(2))

With usual notations, in triangle ABC, a cos (B - C) + b cos (C - A) + c cos (A - B) is equal to

In a triangle ABC, a[b cos C - c cos B]=

In a triangle ABC, a (b cos C - c cos B) =

In a triangle ABC, a (b cos C - c cos B) =

In a triangle ABC, a cos A+b cos B+ c cos C=

In a triangle ABC, a[b cos C - c cos B] =