Answer
$${\text{the improper integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /2} {\tan \theta } d\theta \cr
& \tan \theta {\text{ is not defined at }}x = \pi /2,{\text{so we can write}} \cr
& \int_0^{\pi /2} {\tan \theta } d\theta = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \int_0^b {\tan \theta } d\theta \cr
& {\text{Integrate}} \cr
& \mathop {\lim }\limits_{b \to \pi /{2^ - }} \int_0^b {\tan \theta } d\theta = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \left[ { - \ln \left| {\cos \theta } \right|} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \left[ { - \ln \left| {\cos b} \right| + \ln \left| {\cos 0} \right|} \right] \cr
& = \mathop {\lim }\limits_{b \to \pi /{2^ - }} \left[ { - \ln \left| {\cos b} \right| + \ln \left| 1 \right|} \right] \cr
& = - \mathop {\lim }\limits_{b \to \pi /{2^ - }} \left[ {\ln \left| {\cos b} \right|} \right] \cr
& {\text{Evaluate the limit when }}b \to \pi /{2^ - } \cr
& = - \left[ {\ln \left| {\cos \left( {\frac{\pi }{2}} \right)} \right|} \right] \cr
& = - \infty \cr
& {\text{So}},{\text{ the improper integral diverges}}. \cr
& {\text{The following graph confirms the result}} \cr} $$
