The real value of `theta` for which the expression `(1 - i sin theta)/( 1 + 2 i sin theta)` is purely real is

Options

A. `n pi`
B. `(n + 1) pi// 2 `
C. `(2 n + 1) pi//2`
D. None

Answer

To find the values of \( \theta \) for which the expression

\[ \frac{1 - i \sin \theta}{1 + 2i \sin \theta} \]

is purely real, we will follow these steps:

### Step 1: Identify the expression
Let

\[ z = \frac{1 - i \sin \theta}{1 + 2i \sin \theta} \]

### Step 2: Rationalize the denominator
To simplify, we multiply the numerator and denominator by the conjugate of the denominator:

\[ z = \frac{(1 - i \sin \theta)(1 - 2i \sin \theta)}{(1 + 2i \sin \theta)(1 - 2i \sin \theta)} \]

### Step 3: Simplify the denominator
The denominator simplifies as follows:

\[ (1 + 2i \sin \theta)(1 - 2i \sin \theta) = 1^2 - (2i \sin \theta)^2 = 1 - 4(-1)(\sin^2 \theta) = 1 + 4 \sin^2 \theta \]

### Step 4: Simplify the numerator
Now, simplify the numerator:

\[ (1 - i \sin \theta)(1 - 2i \sin \theta) = 1 - 2i \sin \theta - i \sin \theta + 2(-1)(\sin^2 \theta) = 1 - 3i \sin \theta - 2 \sin^2 \theta \]

### Step 5: Combine the results
Now we can write \( z \) as:

\[ z = \frac{(1 - 2 \sin^2 \theta) - 3i \sin \theta}{1 + 4 \sin^2 \theta} \]

### Step 6: Separate real and imaginary parts
The expression can be separated into real and imaginary parts:

\[ z = \frac{1 - 2 \sin^2 \theta}{1 + 4 \sin^2 \theta} - i \frac{3 \sin \theta}{1 + 4 \sin^2 \theta} \]

### Step 7: Set the imaginary part to zero
For \( z \) to be purely real, the imaginary part must be zero:

\[ - \frac{3 \sin \theta}{1 + 4 \sin^2 \theta} = 0 \]

This implies:

\[ 3 \sin \theta = 0 \]

### Step 8: Solve for \( \theta \)
Thus, we have:

\[ \sin \theta = 0 \]

The solutions for \( \sin \theta = 0 \) are given by:

\[ \theta = n\pi \quad (n \in \mathbb{Z}) \]

### Conclusion
The real values of \( \theta \) for which the expression is purely real are:

\[ \theta = n\pi \]