Answer
negative
Work Step by Step
1. With $-90^{\circ}\lt \theta\lt 90^{\circ}$, then $-270^{\circ}\lt \theta -180^{\circ}\lt -180^{\circ}$
2. This means that $\theta -180^{\circ}$ is in QII or QIII, where $x\lt0$
3. Therefore $\sec( \theta -180)^{\circ}\lt 0$
