$7 \cdot 35$ 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 $A C$, 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 .
(China High School Mathematics League, 1993)
[Solution 1] Draw the right figure, it is obvious that
$$
1=c-a=h=a \text {. }
$$

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

Therefore, the answer is $(A)$.
[Solution 2] $\because c>a, \therefore \angle C>\angle A$.
$$
\begin{array}{l}
\text { Also, } \because \quad h=c-a=\frac{h}{\sin A}-\frac{h}{\sin C}, \\
\therefore \quad \sin C-\sin A=\sin A \cdot \sin C,
\end{array}
$$

That is, $2 \cos \frac{C+A}{2} \sin \frac{C-A}{2}=\frac{1}{2}[\cos (A-C)-\cos (A+C)]$.
However, the right side of the above equation $=\frac{1}{2}\left[1-2 \sin ^{2} \frac{C-A}{2}-2 \cos ^{2} \frac{A+C}{2}+1\right]$
$$
=1-\sin ^{2} \frac{C-A}{2}-\cos ^{2} \frac{C+A}{2} \text {. }
$$

Thus, from $\circledast$ we get $\left(\sin \frac{C-A}{2}+\cos \frac{C+A}{2}\right)^{2}=1$,
which means $\square$
$$
\sin \frac{C-A}{2}+\cos \frac{C+A}{2}= \pm 1 \text {. }
$$

Since $0<\frac{C-A}{2}<\frac{\pi}{2}, 0<\frac{C+A}{2}<\frac{\pi}{2}$,
we have $\sin \frac{C-A}{2}+\cos \frac{C+A}{2}=1$.
Therefore, the answer is $(A)$.