If $\bar{a}$ and $\overline{\mathrm{b}}$ are unit vectors and $\theta$ is the angle between them, then $\tan \theta / 2=$

Given that $\bar{a}$ and $\bar{b}$ are unit vectors, with angle $\theta$ between them, their dot product yields $\bar{a} \cdot \bar{b} = \cos\theta$.

Evaluating $|\bar{a} - \bar{b}|^2$ using vector algebra:

$|\bar{a} - \bar{b}|^2 = 2(1 - \cos\theta) = 4\sin^2(\theta/2)$

Hence, $|\bar{a} - \bar{b}| = 2\sin(\theta/2)$ for $\theta \in [0, \pi]$.

Similarly for $|\bar{a} + \bar{b}|^2$:

$|\bar{a} + \bar{b}|^2 = 2(1 + \cos\theta) = 4\cos^2(\theta/2)$

Thus, $|\bar{a} + \bar{b}| = 2\cos(\theta/2)$.

Taking the ratio yields the tangent of the half-angle:

$\tan(\theta/2) = \frac{|\bar{a} - \bar{b}|}{|\bar{a} + \bar{b}|}$

This result corresponds to option D.