1. **Stating the problem:** We need to find $\cot(30^\circ + \alpha)$ given that $\cot \alpha = 2$. 2. **Formula used:** The cotangent addition formula is $$\cot(A+B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}$$ Here, $A = 30^\circ$ and $B = \alpha$. 3. **Known values:** We know $\cot \alpha = 2$. 4. **Calculate $\cot 30^\circ$:** $$\cot 30^\circ = \sqrt{3}$$ 5. **Apply the formula:** $$\cot(30^\circ + \alpha) = \frac{\cot 30^\circ \cdot \cot \alpha - 1}{\cot 30^\circ + \cot \alpha} = \frac{\sqrt{3} \times 2 - 1}{\sqrt{3} + 2} = \frac{2\sqrt{3} - 1}{\sqrt{3} + 2}$$ 6. **Rationalize the denominator:** Multiply numerator and denominator by the conjugate $\sqrt{3} - 2$: $$\cot(30^\circ + \alpha) = \frac{(2\sqrt{3} - 1)(\sqrt{3} - 2)}{(\sqrt{3} + 2)(\sqrt{3} - 2)}$$ 7. **Simplify denominator:** $$ (\sqrt{3} + 2)(\sqrt{3} - 2) = (\sqrt{3})^2 - 2^2 = 3 - 4 = -1 $$ 8. **Expand numerator:** $$ (2\sqrt{3} - 1)(\sqrt{3} - 2) = 2\sqrt{3} \times \sqrt{3} - 2\sqrt{3} \times 2 - 1 \times \sqrt{3} + 1 \times 2 = 2 \times 3 - 4\sqrt{3} - \sqrt{3} + 2 = 6 - 5\sqrt{3} + 2 = 8 - 5\sqrt{3} $$ 9. **Put it all together:** $$\cot(30^\circ + \alpha) = \frac{8 - 5\sqrt{3}}{-1} = -8 + 5\sqrt{3}$$ **Final answer:** $$\boxed{\cot(30^\circ + \alpha) = -8 + 5\sqrt{3}}$$