Step 1: Recall the triple angle cosine identity.
The standard identity is $\cos 3A = 4\cos^3 A - 3\cos A$. We can factor out $\cos A$ from the right side: $\cos 3A = \cos A(4\cos^2 A - 3)$, which gives \[ 4\cos^2 A - 3 = \frac{\cos 3A}{\cos A}. \] This is the key step because the factors in the problem look exactly like $4\cos^2 A - 3$.
Step 2: Simplify the first factor using the identity.
Set $A = 9^\circ$: \[ 4\cos^2 9^\circ - 3 = \frac{\cos 27^\circ}{\cos 9^\circ}. \]
Step 3: Simplify the second factor using the identity.
Set $A = 27^\circ$: \[ 4\cos^2 27^\circ - 3 = \frac{\cos 81^\circ}{\cos 27^\circ}. \]
Step 4: Multiply the two factors.
\[ (4\cos^2 9^\circ - 3)(4\cos^2 27^\circ - 3) = \frac{\cos 27^\circ}{\cos 9^\circ} \cdot \frac{\cos 81^\circ}{\cos 27^\circ} = \frac{\cos 81^\circ}{\cos 9^\circ}. \] The $\cos 27^\circ$ cancels neatly.
Step 5: Use the complementary angle identity.
Since $81^\circ = 90^\circ - 9^\circ$, we have $\cos 81^\circ = \sin 9^\circ$. Therefore: \[ \frac{\cos 81^\circ}{\cos 9^\circ} = \frac{\sin 9^\circ}{\cos 9^\circ} = \tan 9^\circ. \]
Step 6: State the final answer.
\[ \boxed{\tan 9^\circ} \]