Consider the general term $\frac{\sin \theta }{\cos 3\theta }$. Multiply numerator and denominator by $2\cos \theta$:

$\frac{\sin \theta }{\cos 3\theta }=\frac{2\sin \theta \cos \theta }{2\cos 3\theta \cos \theta }=\frac{\sin 2\theta }{2\cos 3\theta \cos \theta }.$

Write $\sin 2\theta =\sin (3\theta -\theta )=\sin 3\theta \cos \theta -\cos 3\theta \sin \theta$, so

$\frac{\sin \theta }{\cos 3\theta }=\frac{1}{2}\left(\tan 3\theta -\tan \theta \right).$

Apply with $\theta =3^{\circ },9^{\circ },27^{\circ }$:

$A=\frac{1}{2}[(\tan 9^{\circ }-\tan 3^{\circ })+(\tan 27^{\circ }-\tan 9^{\circ })+(\tan 81^{\circ }-\tan 27^{\circ })].$

The sum telescopes to $A=\frac{1}{2}(\tan 81^{\circ }-\tan 3^{\circ })=\frac{B}{2}$.

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