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 $B = \tan 81^\circ - \tan 3^\circ$, find $\frac{B}{A}$

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The expression for $A$ is a sum of terms of the form $\frac{\sin \theta}{\cos 3\theta}$. We seek an identity to convert this into a telescopic series.
Step 2: Key Formula or Approach:
Using the identity: $\tan 3\theta - \tan \theta = \frac{\sin 3\theta \cos \theta - \cos 3\theta \sin \theta}{\cos 3\theta \cos \theta} = \frac{\sin 2\theta}{\cos 3\theta \cos \theta}$.
Multiply numerator and denominator by $\cos \theta$:
$\frac{\sin \theta \cos \theta}{\cos 3\theta \cos \theta} = \frac{\sin 2\theta}{2 \cos 3\theta \cos \theta} = \frac{1}{2}(\tan 3\theta - \tan \theta)$.
So, $\frac{\sin \theta}{\cos 3\theta} = \frac{1}{2}(\tan 3\theta - \tan \theta)$.
Step 3: Detailed Explanation:
1. Apply the identity to each term in $A$:
$T_1 = \frac{\sin 3^\circ}{\cos 9^\circ} = \frac{1}{2}(\tan 9^\circ - \tan 3^\circ)$
$T_2 = \frac{\sin 9^\circ}{\cos 27^\circ} = \frac{1}{2}(\tan 27^\circ - \tan 9^\circ)$
$T_3 = \frac{\sin 27^\circ}{\cos 81^\circ} = \frac{1}{2}(\tan 81^\circ - \tan 27^\circ)$
2. Summing these terms:
$A = \frac{1}{2} [(\tan 9^\circ - \tan 3^\circ) + (\tan 27^\circ - \tan 9^\circ) + (\tan 81^\circ - \tan 27^\circ)]$
$A = \frac{1}{2} (\tan 81^\circ - \tan 3^\circ)$
3. Given $B = \tan 81^\circ - \tan 3^\circ$:
$A = \frac{B}{2} \implies \frac{B}{A} = 2$.
Step 4: Final Answer:
The ratio $B/A$ is 2.