First, check the Assertion:
\[
\cot^{-1}(\sqrt{3}) = \theta.
\text{So,} \cot \theta = \sqrt{3}.
\]
We know,
\[
\cot \frac{\pi}{6} = \frac{\cos \frac{\pi}{6}}{\sin \frac{\pi}{6}}
= \frac{\sqrt{3}/2}{1/2} = \sqrt{3}.
\]
Also, the principal value branch of $\cot^{-1} x$ is $(0, \pi)$, so the angle $\frac{\pi}{6}$ is valid and in principal range.
So, the Assertion is True.
Now, check the Reason:
The domain of the inverse cotangent function is:
\[
\text{Domain}(\cot^{-1} x) = \mathbb{R}, \text{because for any real number } x, \cot^{-1} x \text{ is defined.}
\]
There is no restriction like $\{-1, 1\}$. Maybe the confusion is with $\tan^{-1} x$ where domain is also $\mathbb{R}$. So the Reason is False.
Therefore, Assertion (A) is true but Reason (R) is false.