For any triangle ABC, prove that a+b = cos A-B/2

Do you mean

Mollweide's Formula ?
$(a+b)\sin\frac{C}{2}=c \cdot cos\left(\frac{A-B}{2}\right)$

For any triangle ABC, prove that a+b = cos A-B/2

                                                      2       sin c/2

Do you mean

For any triangle ABC,  prove that

$\frac{a+b}{2}=\frac{cos(A-\frac{B}{2})}{sin(\frac{C}{2})}$