Let a, b, c be the sides of any triangle ABC. Then by applying the sine rule, we get

⇒ a = k sin A, b = k sin B, c = k sin C

So the LHS of the given equation, we get

LHS = a cos A + b cos B + c cos C

Substituting values from sine law, we get

= k sin A cos A + k sin B cos B + k sin C cos C

(∵ π = A + B + C)

Now, from sine rule,

k sin C = c

Putting this value in equation (i), we get

LHS = 2c sin A sin B

And also k sin B = b (from sine rule)

Putting this in equation (i), we get

LHS = 2b sin A sin C

Hence LHS = RHS

i.e., a cos A + b cos B + c cos C = 2b sin A sin C = 2c sin A sin B

Hence proved