Question

If \(A = \frac{\sin 3^\circ}{\cos 9^\circ} + \frac{\sin 9^\circ}{\cos 27^\circ} + \frac{\sin 27^\circ}{\cos 81^\circ}\) and \(B = \tan 81^\circ - \tan 3^\circ\), then \(\frac{B}{A}\) is equal to ____.

Answer 1

Correct Answer: 2

Step 1: Understanding the Concept:
We need to simplify a telescoping trigonometric series.
The terms in sequence \(A\) follow a clear pattern: \(\frac{\sin x}{\cos 3x}\), where \(x\) takes values \(3^\circ, 9^\circ, 27^\circ\).
Step 2: Key Formula or Approach:
We attempt to express the general term \(\frac{\sin x}{\cos 3x}\) as a difference of two terms to create a telescoping sum.
Consider the identity \(\tan 3x - \tan x = \frac{\sin 3x}{\cos 3x} - \frac{\sin x}{\cos x}\).
Simplifying this:
\(\frac{\sin 3x \cos x - \cos 3x \sin x}{\cos x \cos 3x} = \frac{\sin(3x - x)}{\cos x \cos 3x} = \frac{\sin 2x}{\cos x \cos 3x} = \frac{2 \sin x \cos x}{\cos x \cos 3x} = \frac{2 \sin x}{\cos 3x}\).
Thus, we have a beautiful relation: \(\frac{\sin x}{\cos 3x} = \frac{1}{2} (\tan 3x - \tan x)\).
Step 3: Detailed Explanation:
Let's apply this derived identity to each term in the sum \(A\).
For the first term (\(x = 3^\circ\)):
\(\frac{\sin 3^\circ}{\cos 9^\circ} = \frac{1}{2} (\tan 9^\circ - \tan 3^\circ)\).
For the second term (\(x = 9^\circ\)):
\(\frac{\sin 9^\circ}{\cos 27^\circ} = \frac{1}{2} (\tan 27^\circ - \tan 9^\circ)\).
For the third term (\(x = 27^\circ\)):
\(\frac{\sin 27^\circ}{\cos 81^\circ} = \frac{1}{2} (\tan 81^\circ - \tan 27^\circ)\).
Adding these three terms together to find \(A\):
\(A = \frac{1}{2} [(\tan 9^\circ - \tan 3^\circ) + (\tan 27^\circ - \tan 9^\circ) + (\tan 81^\circ - \tan 27^\circ)]\).
Notice that the intermediate terms cancel out (telescoping property):
\(A = \frac{1}{2} (\tan 81^\circ - \tan 3^\circ)\).
Step 4: Final Answer:
The problem defines \(B = \tan 81^\circ - \tan 3^\circ\).
Substitute \(B\) into our expression for \(A\):
\(A = \frac{1}{2} B\).
Rearranging this to find the required ratio \(\frac{B}{A}\):
\(\frac{B}{A} = 2\).