If `(3+2isintheta)/(1- 2isintheta) ` is real , then general value of `theta` is :

Options

A. ` (npi)/(2),nin1`
B. ` npi,nin1`
C. `(npi)/(3),nin1`
D. None of these

Answer

To solve the problem, we need to determine the general value of \(\theta\) such that the expression \(\frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}\) is real.

### Step-by-Step Solution:

1. **Set up the expression**:
We have the expression \(\frac{3 + 2i \sin \theta}{1 - 2i \sin \theta}\).

2. **Rationalize the denominator**:
To eliminate the imaginary part from the denominator, multiply the numerator and the denominator by the conjugate of the denominator:
\[ \frac{(3 + 2i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)} \]

3. **Calculate the denominator**:
The denominator simplifies as follows:
\[ (1 - 2i \sin \theta)(1 + 2i \sin \theta) = 1^2 - (2i \sin \theta)^2 = 1 - 4(-\sin^2 \theta) = 1 + 4 \sin^2 \theta \]

4. **Calculate the numerator**:
The numerator expands to:
\[ (3 + 2i \sin \theta)(1 + 2i \sin \theta) = 3 + 6i \sin \theta + 2i \sin \theta - 4 \sin^2 \theta = 3 - 4 \sin^2 \theta + 8i \sin \theta \]

5. **Combine the results**:
Now we can write the entire expression as:
\[ \frac{3 - 4 \sin^2 \theta + 8i \sin \theta}{1 + 4 \sin^2 \theta} \]

6. **Separate real and imaginary parts**:
The expression can be separated into real and imaginary parts:
- Real part: \(\frac{3 - 4 \sin^2 \theta}{1 + 4 \sin^2 \theta}\)
- Imaginary part: \(\frac{8 \sin \theta}{1 + 4 \sin^2 \theta}\)

7. **Set the imaginary part to zero**:
For the expression to be real, the imaginary part must equal zero:
\[ \frac{8 \sin \theta}{1 + 4 \sin^2 \theta} = 0 \]
This implies that:
\[ 8 \sin \theta = 0 \]

8. **Solve for \(\theta\)**:
From \(8 \sin \theta = 0\), we find:
\[ \sin \theta = 0 \]
The general solution for \(\sin \theta = 0\) is:
\[ \theta = n\pi \quad \text{where } n \text{ is an integer} \]

### Final Answer:
The general value of \(\theta\) is:
\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]