If $$A = \frac{\sin 3°}{\cos 9°} + \frac{\sin 9°}{\cos 27°} + \frac{\sin 27°}{\cos 81°}$$ and $$B = \tan 81° - \tan 3°$$, then $$\frac{B}{A}$$ is equal to _____.

Recall the identity for the difference of two tangents:
$$\tan A-\tan B=\frac{\sin(A-B)}{\cos A\cos B} \quad -(1)$$

Put $$A=3\theta,\;B=\theta$$ in $$(1)$$. Then $$A-B=2\theta$$ and

$$\tan3\theta-\tan\theta=\frac{\sin2\theta}{\cos3\theta\cos\theta} =\frac{2\sin\theta\cos\theta}{\cos3\theta\cos\theta} =\frac{2\sin\theta}{\cos3\theta} \quad -(2)$$

Re-arranging $$(2)$$ we get a very useful relation:

$$\frac{\sin\theta}{\cos3\theta}=\frac12\bigl(\tan3\theta-\tan\theta\bigr) \quad -(3)$$

Now let $$\theta=3^{k}\cdot3^\circ$$ for successive integer values of $$k$$:

Case 0: $$\theta=3^\circ$$ Using $$(3)$$ gives $$\frac{\sin3^\circ}{\cos9^\circ}=\frac12\bigl(\tan9^\circ-\tan3^\circ\bigr) \quad -(4)$$

Case 1: $$\theta=9^\circ$$ $$\frac{\sin9^\circ}{\cos27^\circ}=\frac12\bigl(\tan27^\circ-\tan9^\circ\bigr) \quad -(5)$$

Case 2: $$\theta=27^\circ$$ $$\frac{\sin27^\circ}{\cos81^\circ}=\frac12\bigl(\tan81^\circ-\tan27^\circ\bigr) \quad -(6)$$

Add $$(4),(5),(6)$$ term-by-term. The required sum $$A$$ becomes

$$A=\frac12\Bigl[(\tan9^\circ-\tan3^\circ)+(\tan27^\circ-\tan9^\circ)+(\tan81^\circ-\tan27^\circ)\Bigr]$$

The intermediate terms $$\tan9^\circ$$ and $$\tan27^\circ$$ cancel out, leaving

$$A=\frac12\bigl(\tan81^\circ-\tan3^\circ\bigr) \quad -(7)$$

But $$\tan81^\circ=\tan(90^\circ-9^\circ)=\cot9^\circ$$, so $$(7)$$ can also be written as $$A=\frac12\bigl(\cot9^\circ-\tan3^\circ\bigr).$$

The expression given for $$B$$ is $$B=\tan81^\circ-\tan3^\circ=\cot9^\circ-\tan3^\circ \quad -(8)$$

Comparing $$(7)$$ and $$(8)$$: $$B=2A.$$

Therefore $$\frac{B}{A}=2.$$

Final Answer: 2