in any triangle abc prove

a cosA+b cosB+c cosC=2a sin B sin C

Let ... Δ = area of ΔABC.Then ... bc sin A = ca sin B = ab sin C = 2Δ ... (1)Also ... bc sin A = 2Δ. . . . . sin A = 2Δ / bc.Similarly, ..... sin B = 2Δ / ca ... and ... sin C = 2Δ / ab. ... (2)... LHS

= a cos A + b cos B + c cos C, ... where ... a/sin A = ... = k= (k/2) [ 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ]= (k/2) [ ( sin 2A + sin 2B ) + sin 2C ]= (k/2) [ 2 sin (A+B) cos(A-B) + sin 2C ]= (k/2) [ 2 sin C. cos(A-B) + 2 sin C cos C ]= k sin C [ cos(A-B) + cos C ] ... here .. cos C = cos [ - (A+B) ] = - cos (A+B)= k sin C [ cos(A-B) - cos(A+B) ]= k sin C [ 2 sin A sin B ]= 2 ( k sin A ) sin B sin C= 2 a sin B sin C= Middle Side= 2a [ 2Δ / ca ] [ 2Δ / ab ] ...... from (2)= 8 Δ² / ( abc )= RHS