Five, if $\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}-\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}=2 \tan \alpha$, find the quadrant in which the angle $\alpha$ lies.
Five, Brief Solution: Simplify the given conditions to $\sqrt{\frac{(1+\sin \alpha)^{2}}{\cos ^{2} \alpha}}$
$$
\begin{array}{l}
-\sqrt{\frac{(1-\sin \alpha)^{2}}{\cos ^{2} \alpha}}=2 \operatorname{tg} \alpha, \frac{1+\sin \alpha}{|\operatorname{cos} \alpha|}-\frac{1-\sin \alpha}{|\cos \alpha|} \\
=2 \operatorname{tg} \alpha, \frac{2 \sin \alpha}{|\cos \alpha|}=2 \operatorname{tg} \alpha, \text { since } \operatorname{cos} \alpha>0, \therefore \alpha \in I, N=
\end{array}
$$