3.272. $\frac{\sqrt{\tan \alpha}+\sqrt{\cot \alpha}}{\sqrt{\tan \alpha}-\sqrt{\cot \alpha}}, 0<\alpha<\frac{\pi}{2}$ and $\alpha \neq \frac{\pi}{4}$.
Solution.

$$
\begin{aligned}
& \frac{\sqrt{\tan \alpha}+\sqrt{\cot \alpha}}{\sqrt{\tan \alpha}-\sqrt{\cot \alpha}}=\frac{(\sqrt{\tan \alpha}+\sqrt{\cot \alpha})(\sqrt{\tan \alpha}+\sqrt{\cot \alpha})}{(\sqrt{\tan \alpha}-\sqrt{\cot \alpha})(\sqrt{\tan \alpha}+\sqrt{\cot \alpha})}= \\
& =\frac{(\sqrt{\tan \alpha}+\sqrt{\cot \alpha})^{2}}{(\sqrt{\tan \alpha})^{2}-(\sqrt{\cot \alpha})^{2}}=\frac{\tan \alpha+2+\cot \alpha}{\tan \alpha-\cot \alpha}=\frac{\tan \alpha+2+\frac{1}{\tan \alpha}}{\tan \alpha-\frac{1}{\tan \alpha}}= \\
& =\frac{\tan^{2} \alpha+2 \tan \alpha+1}{\tan^{2} \alpha-1}=\frac{(\tan \alpha+1)^{2}}{(\tan \alpha-1)(\tan \alpha+1)}=\frac{\tan \alpha+1}{\tan \alpha-1}=\frac{\frac{\sin \alpha}{\cos \alpha}+1}{\frac{\sin \alpha}{\cos \alpha}-1}= \\
& =\frac{\sin \alpha+\cos \alpha}{\sin \alpha-\cos \alpha}=\frac{\cos \alpha+\sin \alpha}{\sin \alpha-\cos \alpha}=\frac{\frac{\sqrt{2}}{2} \cos \alpha+\frac{\sqrt{2}}{2} \sin \alpha}{\frac{\sqrt{2}}{2} \sin \alpha-\frac{\sqrt{2}}{2} \cos \alpha}= \\
& =\frac{\cos \alpha \cos \frac{\pi}{4}+\sin \alpha \sin \frac{\pi}{4}}{\sin \alpha \cos \frac{\pi}{4}-\cos \alpha \sin \frac{\pi}{4}}=[\cos x \cos y+\sin x \sin y=\cos (x-y) \\
& \sin x \cos y-\cos x \sin y=\sin (x-y)]=\frac{\cos \left(\alpha-\frac{\pi}{4}\right)}{\sin \left(\alpha-\frac{\pi}{4}\right)}=\cot\left(\alpha-\frac{\pi}{4}\right)
\end{aligned}
$$

Answer: $\cot\left(\alpha-\frac{\pi}{4}\right)$.