\[ \text{Since } \mathbf{a} \text{ and } \mathbf{b} \text{ are unit vectors, we have} \] \[ |\mathbf{a} - \mathbf{b}| = \sqrt{(\mathbf{a} - \mathbf{b})^2} = \sqrt{(\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b})} \] \[ = \sqrt{|\mathbf{a}|^2 + |\mathbf{b}|^2 - 2 \mathbf{a} \cdot \mathbf{b}} = \sqrt{1 + 1 - 2\cos 2\theta} = \sqrt{2(1 - \cos 2\theta)} \] \[ = \sqrt{2(2\sin^2 \theta)} = 2 |\sin \theta| \] \[ \text{Therefore, } |\mathbf{a} - \mathbf{b}| < 1 \text{ implies} \] \[ |\sin \theta| < \frac{1}{2} \Rightarrow \theta \in \left[0, \frac{\pi}{6} \right) \text{ or } \left( \frac{5\pi}{6}, \pi \right] \]