A

Step 1: Calculate \sec 165^\circ. \sec x = \frac{1}{\cos x}, so \sec 165^\circ = \frac{1}{\cos 165^\circ}.
\cos 165^\circ = -\cos(180^\circ - 165^\circ) = -\cos 15^\circ.
\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.
Therefore, \sec 165^\circ = \frac{4}{-\sqrt{6} - \sqrt{2}}.

Step 2: Now calculate \cot 105^\circ. \cot x = \frac{1}{\tan x}, so \cot 105^\circ = \frac{1}{\tan 105^\circ}.
\tan 105^\circ = -\tan(180^\circ - 105^\circ) = -\tan 75^\circ.
Using the angle sum formula, \tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}.
Therefore, \cot 105^\circ = -\frac{1}{\tan 75^\circ}.

Step 3: The value to find is \sec 165^\circ + \cot 105^\circ. Substitute the values obtained above.
Therefore, A.