9.33 Let $\alpha$ be an interior angle of a triangle, and $\alpha \neq$ $90^{\circ}$. Also, $x=\frac{1}{\sin \alpha}-\cos \alpha, y=\sin \alpha-|\operatorname{tg} \alpha|$, then the point $(x, y)$ is in
(A) the first quadrant.
(B) the second quadrant.
(C) the third quadrant.
(D) the fourth quadrant.
(Anhui Province Hefei City Junior High School Mathematics Competition, 1985)
[Solution] Given $01$, and $\cos \alpha0$.
Also, $\because \quad|\operatorname{tg} \alpha|=\left|\frac{\sin \alpha}{\cos \alpha}\right|>|\sin \alpha|=\sin \alpha$,
$\therefore \quad y=\sin \alpha-|\operatorname{tg} \alpha|<0$.
Therefore, $(x, y)$ is in the fourth quadrant.
Hence, the answer is $(D)$.