5. In $\triangle A B C$, the lengths of the sides opposite to angles $A, B, C$ are $a, b, c$ respectively. If $c-a$ equals the height $h$ from vertex $A$ to side $AC$, then the value of $\sin \frac{C-A}{2}+\cos \frac{C+A}{2}$ is
(A) 1;
(B) $\frac{1}{2}$;
(C) $\frac{1}{3}$;
(D) -1.
5. (A)

As shown in the figure,
$$
1=c-a=h=a .
$$

At this moment, $\sin \frac{C-\angle A}{2}+\cos \frac{C+\angle A}{2}$
$$
=\sin 30^{\circ}+\cos 60^{\circ}=1 \text {. }
$$