化简: $\frac{1}{\cos ^{2} \alpha \sqrt{1+\tan ^{2} \alpha}}-\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}(\alpha$ 为第二象限角).
$\because \alpha$ 是第二象限角,

$\therefore \cos \alpha<0$.

$$
\begin{aligned}
& \text { 则原式 }=\frac{1}{\cos ^{2} \alpha \cdot \sqrt{1+\frac{\sin ^{2} \alpha}{\cos ^{2} \alpha}}}-\sqrt{\frac{(1+\sin \alpha)^{2}}{1-\sin ^{2} \alpha}} \\
& =\frac{1}{\cos ^{2} \alpha} \cdot \sqrt{\frac{\cos ^{2} \alpha}{\cos ^{2} \alpha+\sin ^{2} \alpha}}-\frac{1+\sin \alpha}{|\cos \alpha|} \\
& =\frac{-\cos \alpha}{\cos ^{2} \alpha}+\frac{1+\sin \alpha}{\cos \alpha}=\frac{-1+1+\sin \alpha}{\cos \alpha} \\
& =\frac{\sin \alpha}{\cos \alpha}=\tan \alpha .
\end{aligned}
$$