integrate$ \int_{0}^{\pi}\tan(x)\,dx$ there is a discontinutity so I break it at $\pi/2$
$$\begin{align}&\int_{0}^{\pi/2}\tan(x)\,dx+ \int_{\pi}^{\pi/2}\tan(x)\,dx\\ =&[\log(\sec x)]^{\pi/2} _{0} +[\log(\sec x)]^{\pi} _{\pi/2}\\ =&\lim_{x\to\pi/2^{-}}\log(\sec x)+\iota\pi - \lim_{x\to\pi/2^{+}}\log(\sec x)\\ =&\lim_{h\to0}\log(\csc h)+\iota\pi - \lim_{h\to0}\log(-\csc h)\\ =&\lim_{h\to0}\log \left( \frac{\csc h}{\csc h} \right) +\iota\pi - \log(-1)\\ =&\lim_{h\to0}\log \left( \frac{\csc h}{\csc h} \right) +\iota\pi - \iota\pi \\ =&\lim_{h\to0}\log \left( \frac{\csc h}{\csc h} \right) \\ =&\lim_{h\to0}\log(1)\\ =&0\end{align}$$
Is it correct?
Your integral diverges because as you have calculated, you are dealing with $\sec(x)$ which diverges at $\pi/2$ $$[\log(|\sec (x)|)]^{\pi/2} _{0} +[\log(|\sec (x)|)]^{\pi} _{\pi/2}$$
diverges because both limits diverge to infinity.