1. **Stating the problem:** Prove that $$\tan A + 2 \tan^2 A + 4 \tan^4 A + 8 \cot^8 A = \cot A.$$\n\n2. **Recall definitions and identities:** \n- $\tan A = \frac{\sin A}{\cos A}$\n- $\cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A}$\n- Powers of tangent and cotangent follow usual exponent rules.\n\n3. **Rewrite the equation using $\cot A = \frac{1}{\tan A}$:**\n$$\tan A + 2 \tan^2 A + 4 \tan^4 A + 8 \cot^8 A = \tan A + 2 \tan^2 A + 4 \tan^4 A + 8 \left(\frac{1}{\tan A}\right)^8 = \cot A = \frac{1}{\tan A}.$$\n\n4. **Simplify the last term:**\n$$8 \left(\frac{1}{\tan A}\right)^8 = 8 \tan^{-8} A.$$\n\n5. **Rewrite the entire left side as a function of $\tan A = t$:**\n$$t + 2 t^2 + 4 t^4 + 8 t^{-8} = \frac{1}{t}.$$\n\n6. **Multiply both sides by $t^8$ to clear negative powers:**\n$$t^9 + 2 t^{10} + 4 t^{12} + 8 = t^7.$$\n\n7. **Rearrange all terms to one side:**\n$$2 t^{10} + 4 t^{12} + t^9 - t^7 + 8 = 0.$$\n\n8. **Check if this polynomial identity holds for all $t = \tan A$:**\nThis is a complicated polynomial and does not simplify to zero for all $t$.\n\n9. **Conclusion:** The given equation $$\tan A + 2 \tan^2 A + 4 \tan^4 A + 8 \cot^8 A = \cot A$$ is not an identity true for all $A$. It may hold for specific values of $A$ but cannot be proven as a general trigonometric identity.