real analysis - Is function $y=tan x$ uniformly continuous in the open interval $(0,pi/2);?$

Determine whether the function $y=\tan x$ is uniformly continuous in the open interval $(0,\pi/2)$.

I tried approaching it this way

Let $x,y \in (0, \pi/2)$. Then

$$|f(x)-f(y)|=|\tan x-\tan y|={\left|{{\sin x\cos y-\cos x\sin y}\over \cos x\cos y}\right|}\le|\sin(x-y)|\le|x-y|$$

Selecting $\delta=\epsilon$ we have that the given function is uniformly continuous.

Where am i gong wrong ?

Answer

Every uniformly continuous function is bounded on every bounded interval included in its domain of definition. The proof only uses that the real line is Archimedean.