Ratio of Trigonometric Expressions with Tangent and Secant Terms

To find the value of the ratio $\frac{B}{A}$, we first simplify the expression for $A$.

Consider the general term in the expression for $A$, which is of the form: $T(x) = \frac{\sin x}{\cos 3x}$

Multiply and divide $T(x)$ by $2\cos x$: $T(x) = \frac{2\sin x \cos x}{2\cos 3x \cos x} = \frac{\sin 2x}{2\cos 3x \cos x}$

Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$ for $A = 3x$ and $B = x$, we have: $\sin 2x = \sin(3x - x) = \sin 3x \cos x - \cos 3x \sin x$

Substituting this into the expression for $T(x)$: $T(x) = \frac{\sin 3x \cos x - \cos 3x \sin x}{2\cos 3x \cos x}$ $T(x) = \frac{1}{2} \left( \frac{\sin 3x \cos x}{\cos 3x \cos x} - \frac{\cos 3x \sin x}{\cos 3x \cos x} \right)$ $T(x) = \frac{1}{2} (\tan 3x - \tan x)$

Now, applying this identity to each term of $A$:

1. For $x = 3^\circ$: $\frac{\sin 3^\circ}{\cos 9^\circ} = \frac{1}{2} (\tan 9^\circ - \tan 3^\circ)$

2. For $x = 9^\circ$: $\frac{\sin 9^\circ}{\cos 27^\circ} = \frac{1}{2} (\tan 27^\circ - \tan 9^\circ)$

3. For $x = 27^\circ$: $\frac{\sin 27^\circ}{\cos 81^\circ} = \frac{1}{2} (\tan 81^\circ - \tan 27^\circ)$

Summing these three terms gives: $A = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}$ $A = \frac{1}{2} \left[ (\tan 9^\circ - \tan 3^\circ) + (\tan 27^\circ - \tan 9^\circ) + (\tan 81^\circ - \tan 27^\circ) \right]$

Since this is a telescoping sum, intermediate terms cancel out: $A = \frac{1}{2} (\tan 81^\circ - \tan 3^\circ)$

We are given that $B = \tan 81^\circ - \tan 3^\circ$. Therefore: $A = \frac{1}{2} B \implies \frac{B}{A} = 2$