3.278. $\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}-\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}$, if: a) $90^{\circ}<\alpha<180^{\circ}$;

b) $270^{\circ}<\alpha<360^{\circ}$.
Solution.

$\sqrt{\frac{1+\sin \alpha}{1-\sin \alpha}}-\sqrt{\frac{1-\sin \alpha}{1+\sin \alpha}}=\frac{(\sqrt{1+\sin \alpha})^{2}-(\sqrt{1-\sin \alpha})^{2}}{\sqrt{(1-\sin \alpha)(1+\sin \alpha)}}=$

$=\frac{1+\sin \alpha-1+\sin \alpha}{\sqrt{1-\sin ^{2} \alpha}}=\frac{2 \sin \alpha}{\sqrt{\cos ^{2} \alpha}}=\frac{2 \sin \alpha}{|\cos \alpha|}=$

$=\left\{\begin{array}{l}-2 \tan \alpha, \text { if } 90^{\circ}<\alpha<180^{\circ} ; \\ 2 \tan \alpha, \text { if } 270^{\circ}<\alpha<360^{\circ} .\end{array}\right.$

Answer: a) $-2 \tan \alpha ;$ b) $2 \tan \alpha$.