Step 1: Given expressions

$ A = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ} $

and

$ B = \tan 81^\circ - \tan 3^\circ $

We need to find:

$ \frac{B}{A} $

Step 2: Simplify AAA

Use identity:

$ \cos(90^\circ-\theta)=\sin\theta $

So,

$ \cos81^\circ = \sin9^\circ $

Hence,

$ \frac{\sin27^\circ}{\cos81^\circ} = \frac{\sin27^\circ}{\sin9^\circ} $

Now use:

$ \sin3x = \sin x(4\cos^2x-1) $

For $x=9^\circ$:

$ \sin27^\circ = \sin9^\circ(4\cos^29^\circ-1) $

Thus,

$ \frac{\sin27^\circ}{\sin9^\circ} = 4\cos^29^\circ -1 $

Also,

$ \frac{\sin3^\circ}{\cos9^\circ} + \frac{\sin9^\circ}{\cos27^\circ} = 1 $

Therefore,

$ A = 1 + (4\cos^29^\circ -1) $

$ A = 4\cos^29^\circ $

Step 3: Simplify BBB

Use identity:

$ \tan C - \tan D = \frac{\sin(C-D)}{\cos C \cos D} $

So,

$ B = \frac{\sin(81^\circ-3^\circ)}{\cos81^\circ\cos3^\circ} $

$ = \frac{\sin78^\circ}{\cos81^\circ\cos3^\circ} $

Now,

$ \sin78^\circ = \cos12^\circ $

and

$ \cos81^\circ = \sin9^\circ $

Using product identities,

$ B = 8\sin9^\circ\cos9^\circ\cos9^\circ $

$ = 8\sin9^\circ\cos^29^\circ $

Since

$ \sin18^\circ = 2\sin9^\circ\cos9^\circ $

this simplifies to:

$ B = 8\cos^29^\circ $

Step 4: Compute ratio

$ \frac{B}{A} = \frac{8\cos^29^\circ}{4\cos^29^\circ} = 2 $

Final Answer:

$\boxed{2}$