Since any function which is continuous on a closed and bounded interval is uniformly continuous, in that way the given function is uniformly continuous.
And in the $\epsilon, \delta$ method I tried this much. For $x,y\in[0,\pi/4]$ $|x-y|\leq \pi/4(=\delta) \implies|f(x)-f(y)|=|\tan x-\tan y|=|\frac{\sin(x-y)}{\cos x.\cos y}|\leq 1=\epsilon$ Then choosing $\delta=\pi\epsilon/4$ , we have uniformly contnuity of the function. Should I do anything more?
Best Answer
Your proof is not correct, since it relies on the choice $\epsilon = 1$; you need to be much more precise with the estimates. Starting at your end,
\begin{align*} \left|\frac{\sin{(x - y)}}{\cos x \cos y}\right| &\le \frac{|x - y|}{\left(\cos \pi/4\right)^2} \end{align*}
where we have used that $|\sin(z)| \le |z|$ and that cosine is decreasing on $[0, \pi/4]$. This should give the choice of $\delta$, as well as show you why $\tan$ is not uniformly continuous on, say, $[0, \pi/2]$.