Answer
See proof below.
Work Step by Step
$r(\theta) $ is continuous and differentiable on $(0,\displaystyle \frac{\pi}{2})$.
$r(\displaystyle \frac{\pi}{6})\approx-1.678\lt 0, \qquad r(\frac{\pi}{3})\approx 0.108\gt 0$
so the Intermediate Value Theorem guarantees that there is a zero between $\displaystyle \frac{\pi}{6}$ and $\displaystyle \frac{\pi}{3}$.
Check whether $r$ is increasing over $(0,\displaystyle \frac{\pi}{2})$.
(If it is, the mentioned zero would be the only zero.)
$r'(\theta)=\sec^{2}\theta+\csc^{2}\theta-1\geq 1+1-1\gt 0$
$r'$ is always positive on$ (0,\displaystyle \frac{\pi}{2}),$ which means that $r $ is an increasing function on$ (0,\displaystyle \frac{\pi}{2})$
So, $r$ has only one zero in $(0,\displaystyle \frac{\pi}{2})$
