1. **State the problem:** Find the exact value of $\cot(-510^\circ)$.\n\n2. **Recall the cotangent function and angle properties:** \nThe cotangent function is defined as $\cot \theta = \frac{\cos \theta}{\sin \theta}$.\nAlso, cotangent is periodic with period $180^\circ$, so $\cot(\theta) = \cot(\theta + 180^\circ k)$ for any integer $k$.\n\n3. **Reduce the angle $-510^\circ$ to an equivalent angle between $0^\circ$ and $360^\circ$: **\nAdd $360^\circ$ repeatedly until the angle is positive: \n$$-510^\circ + 360^\circ = -150^\circ$$\nStill negative, add $360^\circ$ again: \n$$-150^\circ + 360^\circ = 210^\circ$$\nSo, $\cot(-510^\circ) = \cot(210^\circ)$.\n\n4. **Use periodicity of cotangent with period $180^\circ$: **\n$$\cot(210^\circ) = \cot(210^\circ - 180^\circ) = \cot(30^\circ)$$\n\n5. **Evaluate $\cot(30^\circ)$: **\nRecall $\sin 30^\circ = \frac{1}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$.\nTherefore, \n$$\cot 30^\circ = \frac{\cos 30^\circ}{\sin 30^\circ} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$$\n\n**Final answer:** \n$$\boxed{\cot(-510^\circ) = \sqrt{3}}$$