9. For what values of $\alpha$ is the equality

$$
\sqrt{\operatorname{tg}^{2} \alpha-\sin ^{2} \alpha}=\operatorname{tg} \alpha \cdot \sin \alpha ?
$$
9. Transforming the left side into the square of the right, we get $\sqrt{\operatorname{tg}^{2} \alpha-\sin ^{2} \alpha}=|\operatorname{tg} \alpha \sin \alpha|$. In this case, $|\operatorname{tg} \alpha \sin \alpha|=\operatorname{tg} \alpha \sin \alpha$. This equality is valid under the condition that $\operatorname{tg} \alpha \sin \alpha>0$, i.e., $\cos \alpha>0$, from which $-\frac{\pi}{2}+2 \pi k<\alpha<\frac{\pi}{2}+2 \pi k$ or $(4 k-1) \frac{\pi}{2}<\alpha<(4 k+1) \frac{\pi}{2} ; \quad k=0$; $\pm 1, \pm 2, \ldots$