1. **State the problem:** Find the exact value of $\cot(-510^\circ)$.\n\n2. **Recall the cotangent function and angle reduction:** The cotangent function is periodic with period $180^\circ$, so $\cot(\theta) = \cot(\theta + 180^\circ k)$ for any integer $k$. Also, $\cot(-\theta) = -\cot(\theta)$.\n\n3. **Reduce the angle $-510^\circ$ to an equivalent angle between $0^\circ$ and $360^\circ$:}\n$$-510^\circ + 3 \times 180^\circ = -510^\circ + 540^\circ = 30^\circ$$\nSo, $\cot(-510^\circ) = \cot(30^\circ)$.\n\n4. **Use the odd function property:**\n$$\cot(-510^\circ) = -\cot(510^\circ)$$\nBut since $\cot(510^\circ) = \cot(510^\circ - 360^\circ) = \cot(150^\circ)$, we can also write:\n$$\cot(-510^\circ) = -\cot(150^\circ)$$\n\n5. **Evaluate $\cot(150^\circ)$:**\n$150^\circ = 180^\circ - 30^\circ$, and $\cot(180^\circ - \theta) = -\cot(\theta)$, so:\n$$\cot(150^\circ) = -\cot(30^\circ)$$\n\n6. **Evaluate $\cot(30^\circ)$:**\n$$\cot(30^\circ) = \frac{\cos 30^\circ}{\sin 30^\circ} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$$\n\n7. **Putting it all together:**\n$$\cot(-510^\circ) = -\cot(150^\circ) = -(-\cot(30^\circ)) = \cot(30^\circ) = \sqrt{3}$$\n\n**Final answer:** $\boxed{\sqrt{3}}$