Question

If \( 1 + \sin\theta + \sin^2\theta + \dots \text{ upto } \infty = 2\sqrt{3} + 4 \), then \( \theta = \)

• A. \( \frac{3\pi}{4} \)
• B. \( \frac{\pi}{3} \)
• C. \( \frac{\pi}{4} \)
• D. \( \frac{\pi}{6} \)

Hint:

To save time during the exam, you can substitute the options directly:
- For \( \theta = \frac{\pi}{3} \), \( \sin\theta = \frac{\sqrt{3}}{2} \).
- Sum \( S_\infty = \frac{1}{1 - \sqrt{3}/2} = \frac{2}{2 - \sqrt{3}} = 2(2 + \sqrt{3}) = 4 + 2\sqrt{3} \), which matches perfectly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to solve the given infinite series equation involving trigonometric terms to find the value of \( \theta \).

Step 2: Key Formula or Approach:
The sum of an infinite geometric progression (G.P.) with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:
\[ S_\infty = \frac{a}{1 - r} \]

Step 3: Detailed Explanation:
The given infinite series is:
\[ 1 + \sin\theta + \sin^2\theta + \dots = 2\sqrt{3} + 4 \] This is an infinite G.P. where:
- First term, \( a = 1 \)
- Common ratio, \( r = \sin\theta \)
Using the sum formula:
\[ \frac{1}{1 - \sin\theta} = 2\sqrt{3} + 4 \] Factorize the right-hand side:
\[ \frac{1}{1 - \sin\theta} = 2(2 + \sqrt{3}) \] Taking reciprocal on both sides:
\[ 1 - \sin\theta = \frac{1}{2(2 + \sqrt{3})} \] Rationalize the denominator by multiplying the numerator and denominator by \( (2 - \sqrt{3}) \):
\[ 1 - \sin\theta = \frac{2 - \sqrt{3}}{2(2 + \sqrt{3})(2 - \sqrt{3})} \] \[ 1 - \sin\theta = \frac{2 - \sqrt{3}}{2(4 - 3)} = \frac{2 - \sqrt{3}}{2} = 1 - \frac{\sqrt{3}}{2} \] Comparing both sides:
\[ \sin\theta = \frac{\sqrt{3}}{2} \] Since \( \theta \) lies in the standard domain of the options:
\[ \theta = \frac{\pi}{3} \]

Step 4: Final Answer:
(B) \( \frac{\pi}{3} \)