Answer
See graph and explanations.
Work Step by Step
Step 1. Given the first derivative $y'=tan^2\theta-1=sec^2\theta-2, -\frac{\pi}{2}\lt\theta\lt\frac{\pi}{2}$, we have $y''=2sec^2\theta\ tan\theta$
Step 2. Possible critical points are $y'=0, \theta=\pm\frac{\pi}{4}$. Determine the increasing $y'\gt0$ and decreasing $y'\lt0$ regions by signs of $y'$: $(-\frac{\pi}{2})..(+)..(-\frac{\pi}{4})..(-)..(\frac{\pi}{4})..(+)..(\frac{\pi}{2})$; thus the function increases on $(-\frac{\pi}{2},-\frac{\pi}{4}), (\frac{\pi}{4}, \frac{\pi}{2})$, decreases on $(-\frac{\pi}{4}, \frac{\pi}{4})$. Local maximum at $\theta=-\frac{\pi}{4}$ and local minimum at $\theta=\frac{\pi}{4}$.
Step 3. Use signs of $y''$ to determine concavity and possible inflection points at $\theta=0$: $(-\frac{\pi}{2})..(-)..(0)..(+)..(\frac{\pi}{2})$. Thus the function is concave down on $(-\frac{\pi}{2},0)$ and concave up on $(0,\frac{\pi}{2})$ with $\theta=\pi$ as the inflection point.
Step 4. Sketch the function based on the above results as shown.
