D

Step 1: Given that \( \tan \theta = \sqrt{3} \), we can find the value of \( \theta \). Since \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we know that this corresponds to an angle of \( 60^{\circ} \) (or equivalently \( \frac{\pi}{3} \) radians) in the first quadrant.
Step 2: From the right triangle properties, we have:
\( \sin \theta = \frac{\sqrt{3}}{2} \) and \( \cos \theta = \frac{1}{2} \).
Step 3: Now, we need to find the expression: \( sin \theta + cos \theta \) and check its value.
\( \sin \theta + \cos \theta = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \)
Step 4: Since we need to consider that the sum might be negative, we should consider negative angles. The angles where \( \tan \) is positive and \( \sin \) and \( \cos \) are negative (i.e., in the 3rd quadrant) would be relevant; hence, \( \theta = 240^{\circ} \) (or \( \frac{4\pi}{3} \)).
Final Step: Looking at the angles in the 3rd quadrant, we find:
\( \sin(240^{\circ}) = -\frac{\sqrt{3}}{2} \) and \( \cos(240^{\circ}) = -\frac{1}{2} \). Thus the sum \( \sin 240^{\circ} + \cos 240^{\circ} = -\frac{\sqrt{3} + 1}{2} < 0 \).
Therefore the required value is \( sin \theta + cos \theta < 0 \). Hence, option D matches the conditions set forth in the question.