Answer
See explanations.
Work Step by Step
a. Given $y=tan\theta$, we have $y'=sec^2\theta\gt0$. Thus the function increases on every open interval in its domain.
b. The function is periodic and has open domain intervals $(k\pi-\frac{\pi}{2}, k\pi+\frac{\pi}{2})$, where $k$ is an integer. We can only compare function values when they fall in the same domain interval. In this case, $x=\frac{\pi}{4}$ and $x=\pi$ are not in the same interval and there is an asymptote at $x=\frac{\pi}{2}$ in between, as shown in the graph.
