\(\frac{{\cos A + \cos B}}{{a + b}} = \frac{{\left( {b + c - a} \right)\left( {c + a - b} \right)}}{{2abc}}\)

⇔ cosA + cosB = \(\frac{{\left( {a + b} \right)\left[ {\left( {c + \left( {b - a} \right)} \right)\left( {c - \left( {b - a} \right)} \right)} \right]}}{{2abc}}\)

⇔ \(\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + \frac{{{c^2} + {a^2} - {b^2}}}{{2ac}} = \frac{{\left( {a + b} \right)\left[ {{c^2} - {{\left( {b - a} \right)}^2}} \right]}}{{2abc}}\)

⇔ ab² + ac² – a³ + c²b + a²b – b³ = (a + b)(c² + 2ab – b² – a²)

⇔ ab² + ac² – a³ + c²b + a²b – b³ = ac² + 2a²b – b²a – a³ + 2ab² – b³ – ba²

⇔ 0 = a²b – b²a + ab² – ba²

⇔ 0 = 0 (đúng)

Vậy \(\frac{{\cos A + \cos B}}{{a + b}} = \frac{{\left( {b + c - a} \right)\left( {c + a - b} \right)}}{{2abc}}\).