2. If $0<x<\frac{\pi}{2}$, then the range of $\sec x+\csc x$ is

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2. $2 \sqrt{2},+\infty$ )

Let $t=\sin x+\cos x, x \in\left(0, \frac{\pi}{2}\right)$. Then $t=\sqrt{2} \sin \left(x+\frac{\pi}{4}\right) \in(1, \sqrt{2}]$, $\sec x+\csc x=\frac{1}{\sin x}+\frac{1}{\cos x}=\frac{\sin x+\cos x}{\sin x \cos x}=\frac{t}{\frac{t^{2}-1}{2}}=\frac{2}{t-\frac{1}{t}}$. $t \in(1, \sqrt{2}]$ then $t-\frac{1}{t} \in\left(0, \frac{\sqrt{2}}{2}\right]$, so $\sec x+\csc x \in[2 \sqrt{2},+\infty)$.