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 $\left(\cos \frac{A}{2}-\sin \frac{A}{2}\right) \cdot\left(\sin \frac{C}{2}+\cos \frac{C}{2}\right)=$ $\qquad$ .
1781 . Since $c-a=h$, $A$ is an acute angle, $\cos \frac{A}{2}>\sin \frac{A}{2}$. And $c-a=c \sin A$, that is
$$
\begin{aligned}
& \sin C-\sin A=\sin C \sin A . \\
& \left(\cos \frac{A}{2}-\sin \frac{A}{2}\right)^{2}\left(\sin \frac{C}{2}+\cos \frac{C}{2}\right)^{2} \\
= & (1-\sin A)(1+\sin C) \\
= & 1-\sin A+\sin C-\sin A \sin C \\
= & 1 .
\end{aligned}
$$

Therefore,
$$
\left(\cos \frac{A}{2}-\sin \frac{A}{2}\right)\left(\sin \frac{C}{2}+\cos \frac{C}{2}\right)=1
$$