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

Here we will consider LHS, so we get

LHS = b cos B + c cos C

Substituting corresponding values from sine rule in above equation, we get

⇒ = k sin B cos B + k sin C cos C

Now we will consider RHS, so we get

RHS = a cos (B – C)

Substituting corresponding values from sine rule in the above equation, we get

⇒ = k sin A cos (B – C)

But sin(A + B) + sin (A - B) = 2 sin A cos B, so the above equation becomes,

We know in a triangle, π = A + B + C, hence the above equation becomes,

Comparing equation (i) and (ii),

LHS = RHS

Hence proved