Given,

      a cos A = b cos B

=> k * sin A * cos A = k * sin B * cos B             {by sin formula}

=> sin A * cos A = sin B * cos B

=> 2 * sin A * cos A = 2 * sin B * cos B

=> sin 2A = cos 2B               {sin 2A = 2 * sin A * cos A}

=> sin 2A - cos 2B = 0

=> 2 * cos(A + B) * sin(A - B) = 0       {Apply sin C - sin D formula}

Now, either cos(A + B) = 0

=> cos(A + B) = cos 90

=> A + B = 90 degree

So, C = 90 degree

So, triangle ABC is rigth angled at C.

Again, sin(A - B) = 0

=> sin(A - B) = sin 0

=> A - B = 0

=> A = B

So, triangle ABC is isosceles.

Hence, if a cos A = b cos B then either triangle ABC is right angled at C or triangle ABC is isosceles.