Here's a nice little graphical solution, (which I believe is what your observation was): notice that the graph of $f(x)=\lfloor{\tan(x)} \rfloor$ has a sort of symmetry such that (for non-integers), we have $f(x)+f(-x)=-1$. Thus we can write the integral as $$\int_{-\pi/2}^{\pi/2} f(x)\ dx=\int_0^{\pi/2} f(x)+f(-x) \ dx = \int_0^{\pi/2} -1 \ dx = \color{red}{-\pi/2}$$
I'll add a picture here to show the observation that is guiding my equivalence of integrals. Also note that the last integral could really just the formula for an area of a rectangle, so I would definitely call this a geometric approach.

On a cooler note, you can actually do this with any odd function (as long as the function doesn't take on the integers in any sizeable domain) as in general for $y$ is not an integer, we have
$$\lfloor y \rfloor + \lfloor -y \rfloor= -1$$
Edit: To answer your question about the ceiling function, try to do the same type of thing based on the observation that the following holds when $y$ is not an integer
$$\lceil y \rceil + \lceil -y \rceil = 1$$