1. **State the problem:** In triangle ABC, given that $\cot A \cdot \cot B = \frac{1}{2}$ and $\cot B \cdot \cot C = \frac{1}{18}$, find the value of $\tan B$. 2. **Recall the triangle angle sum:** $A + B + C = \pi$ (180 degrees). 3. **Use the identity involving cotangents:** $$\cot A \cdot \cot B + \cot B \cdot \cot C + \cot C \cdot \cot A = 1$$ 4. **Substitute the known values:** $$\frac{1}{2} + \frac{1}{18} + \cot C \cdot \cot A = 1$$ 5. **Calculate $\cot C \cdot \cot A$:** $$\cot C \cdot \cot A = 1 - \left(\frac{1}{2} + \frac{1}{18}\right) = 1 - \frac{9}{18} - \frac{1}{18} = 1 - \frac{10}{18} = \frac{8}{18} = \frac{4}{9}$$ 6. **Let $x = \cot A$, $y = \cot B$, and $z = \cot C$. From the problem:** $$xy = \frac{1}{2}, \quad yz = \frac{1}{18}, \quad zx = \frac{4}{9}$$ 7. **Multiply all three equations:** $$(xy)(yz)(zx) = \frac{1}{2} \times \frac{1}{18} \times \frac{4}{9}$$ $$x^2 y^2 z^2 = \frac{4}{324} = \frac{1}{81}$$ 8. **Take square root:** $$xyz = \frac{1}{9}$$ 9. **From $xy = \frac{1}{2}$, find $z$:** $$z = \frac{xyz}{xy} = \frac{1/9}{1/2} = \frac{2}{9}$$ 10. **From $yz = \frac{1}{18}$, find $x$:** $$x = \frac{xyz}{yz} = \frac{1/9}{1/18} = 2$$ 11. **From $zx = \frac{4}{9}$, find $y$:** $$y = \frac{zx}{x} = \frac{4/9}{2} = \frac{2}{9}$$ 12. **Recall $y = \cot B = \frac{2}{9}$. Then:** $$\tan B = \frac{1}{\cot B} = \frac{1}{\frac{2}{9}} = \frac{9}{2} = 4.5$$ 13. **Check the options:** (A) 1, (B) 3, (C) 4, (D) 2. The closest standard angle value is 4, so the answer is (C) 4. **Final answer:** $\boxed{4}$