1. **Stating the problem:** Given that $\tan(a) + \tan(b) + \tan(c) = \cot(a) + \cot(b) + \cot(c)$ where $a, b,$ and $c$ are angles of a triangle. 2. **Recall the triangle angle sum:** Since $a, b,$ and $c$ are angles of a triangle, we have $$a + b + c = 180^\circ.$$ 3. **Rewrite cotangent in terms of tangent:** Recall that $\cot(\theta) = \frac{1}{\tan(\theta)}$. 4. **Rewrite the equation:** $$\tan(a) + \tan(b) + \tan(c) = \frac{1}{\tan(a)} + \frac{1}{\tan(b)} + \frac{1}{\tan(c)}.$$ 5. **Multiply both sides by $\tan(a)\tan(b)\tan(c)$ to clear denominators:** $$\tan(a)\tan(b)\tan(c) \left( \tan(a) + \tan(b) + \tan(c) \right) = \tan(a)\tan(b)\tan(c) \left( \frac{1}{\tan(a)} + \frac{1}{\tan(b)} + \frac{1}{\tan(c)} \right).$$ 6. **Simplify each side:** Left side: $$\tan(a)\tan(b)\tan(c) \left( \tan(a) + \tan(b) + \tan(c) \right) = \tan^2(a)\tan(b)\tan(c) + \tan(a)\tan^2(b)\tan(c) + \tan(a)\tan(b)\tan^2(c).$$ Right side: $$\tan(a)\tan(b)\tan(c) \left( \frac{1}{\tan(a)} + \frac{1}{\tan(b)} + \frac{1}{\tan(c)} \right) = \tan(b)\tan(c) + \tan(a)\tan(c) + \tan(a)\tan(b).$$ 7. **Set the equation:** $$\tan^2(a)\tan(b)\tan(c) + \tan(a)\tan^2(b)\tan(c) + \tan(a)\tan(b)\tan^2(c) = \tan(b)\tan(c) + \tan(a)\tan(c) + \tan(a)\tan(b).$$ 8. **Let $x = \tan(a), y = \tan(b), z = \tan(c)$ for simplicity:** $$x^2 y z + x y^2 z + x y z^2 = y z + x z + x y.$$ 9. **Factor the left side:** $$x y z (x + y + z) = y z + x z + x y.$$ 10. **Recall the tangent sum formula for a triangle:** Since $a + b + c = 180^\circ$, we have $$\tan(a + b + c) = \tan(180^\circ) = 0.$$ Using the tangent addition formula: $$\tan(a + b + c) = \frac{\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c)}{1 - (\tan(a)\tan(b) + \tan(b)\tan(c) + \tan(c)\tan(a))} = 0.$$ 11. **Set numerator to zero (since denominator cannot be zero):** $$\tan(a) + \tan(b) + \tan(c) - \tan(a)\tan(b)\tan(c) = 0,$$ which means $$x + y + z = x y z.$$ 12. **Compare with step 9:** From step 9, $$x y z (x + y + z) = y z + x z + x y,$$ substitute $x + y + z = x y z$: $$x y z \cdot x y z = y z + x z + x y,$$ which is $$ (x y z)^2 = y z + x z + x y.$$ 13. **Rewrite:** $$ (x y z)^2 = y z + x z + x y.$$ 14. **Conclusion:** The given equation implies the relation $$x + y + z = x y z$$ where $x = \tan(a), y = \tan(b), z = \tan(c)$ and $a, b, c$ are angles of a triangle. **Final answer:** $$\boxed{\tan(a) + \tan(b) + \tan(c) = \tan(a) \tan(b) \tan(c)}.$$