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

Explanation

`acos(B-C)+bcos(C-A)+c cos(A-B)`

`=2RsinAcos(B-C)+2RsinBcos(C-A)+2RsinC cos(A-B)`

`=2Rsin(pi-(B+C))cos(B-C)+2Rsin(pi-(C+A))cos(C-A)+2Rsin(pi-(A+B)) cos(A-B)`

`=2Rsin(B+C)cos(B-C)+2Rsin(C+A)cos(C-A)+2Rsin(A+B) cos(A-B)`

`=R(sin2B+sin2C+sin2C+sin2A+sin2A+sin2B)`

`=2R(sin2A+sin2B+sin2C)`

`=2R(4sinAsinBsinC)`

`= 8RsinAsinBsinC`

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