$26 \cdot 13$ Let $r$ be the distance from the origin to point $P$, given that the coordinates of point $P$ are $(x, y)$. If $s=\frac{y}{r}, c=\frac{x}{r}$, then the range of $s^{2}-c^{2}$ is
(A) $(-\infty,-1)$ or $(1,+\infty)$.
(B) $(-\infty,-1]$ or $[1,+\infty)$.
(C) $(-1,1)$.
(D) $[-1,1]$.
(E) only -1 and +1.
(9th American High School Mathematics Examination, 1958)
[Solution] Given $r=\sqrt{x^{2}+y^{2}}$, therefore
$$
s^{2}-c^{2}=\left(\frac{y}{r}\right)^{2}-\left(\frac{x}{r}\right)^{2}=\frac{y^{2}-x^{2}}{x^{2}+y^{2}}=1-\frac{2 x^{2}}{x^{2}+y^{2}} \text {. }
$$

Also, $\frac{-2 x^{2}}{x^{2}+y^{2}}=\left\{\begin{array}{l}0, \text { when } x=0, y=r \text {; } \\ 2, \text { when } x=r, y=0 \text { . }\end{array}\right.$
Thus, $\quad-1 \leqslant 1-\frac{2 x^{2}}{x^{2}+y^{2}} \leqslant 1$,
which means $\square$
$$
-1 \leqslant s^{2}-c^{2} \leqslant 1 .
$$

Therefore, the answer is $(D)$.