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:
Each term in $A$ is of the form $\frac{\sin \theta}{\cos 3\theta}$. We can use the identity $\frac{\sin \theta}{\cos 3\theta} = \frac{1}{2}(\tan 3\theta - \tan \theta)$ to simplify the summation.

Step 2: Key Formula or Approach:
Identity: $\tan 3\theta - \tan \theta = \frac{\sin(3\theta - \theta)}{\cos 3\theta \cos \theta} = \frac{\sin 2\theta}{\cos 3\theta \cos \theta} = \frac{2\sin \theta \cos \theta}{\cos 3\theta \cos \theta} = \frac{2\sin \theta}{\cos 3\theta}$. Therefore, $\frac{\sin \theta}{\cos 3\theta} = \frac{1}{2}(\tan 3\theta - \tan \theta)$.

Step 3: Detailed Explanation:
1. Let $\theta = 3^\circ, 9^\circ, 27^\circ$.
2. $A = \frac{1}{2} [ (\tan 9^\circ - \tan 3^\circ) + (\tan 27^\circ - \tan 9^\circ) + (\tan 81^\circ - \tan 27^\circ) ]$.
3. Notice the telescoping sum: intermediate terms cancel out.
4. $A = \frac{1}{2} (\tan 81^\circ - \tan 3^\circ)$.
5. Since $B = \tan 81^\circ - \tan 3^\circ$, we have $A = \frac{1}{2}B$.
6. $\frac{B}{A} = \frac{B}{B/2} = 2$.

Step 4: Final Answer:
The value of $\frac{B}{A}$ is 2.