178 In $\triangle A B C$, the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $c-a$ equals the height $h$ from $A C$, then $\left(\cos \frac{A}{2}-\sin \frac{A}{2}\right) \cdot\left(\sin \frac{C}{2}+\cos \frac{C}{2}\right)=$ $\qquad$ .