The real value of `alpha` for which the expression `(1- isin alpha)/(1+2i sinalpha)` is purely real is

Options

A. `(n+1) (pi)/(2) n , in N`
B. `(2n+1) (pi)/(2) n , in N`
C. `n pi, n in N`
D. none of these

Answer

To find the real value of \( \alpha \) for which the expression

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

is purely real, we can follow these steps:

### Step 1: Rationalize the denominator

We multiply the numerator and the denominator by the conjugate of the denominator:

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

### Step 2: Simplify the numerator

The numerator becomes:

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

Since \( i^2 = -1 \), we have:

\[ = 1 - 2i \sin \alpha - i \sin \alpha - 2 \sin^2 \alpha = 1 - 2 \sin^2 \alpha - 3i \sin \alpha \]

### Step 3: Simplify the denominator

The denominator is:

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

### Step 4: Combine the results

Now we can rewrite the expression:

\[ \frac{1 - 2 \sin^2 \alpha - 3i \sin \alpha}{1 + 4 \sin^2 \alpha} \]

### Step 5: Separate real and imaginary parts

This can be expressed as:

\[ \frac{1 - 2 \sin^2 \alpha}{1 + 4 \sin^2 \alpha} - \frac{3i \sin \alpha}{1 + 4 \sin^2 \alpha} \]

### Step 6: Set the imaginary part to zero

For the expression to be purely real, the imaginary part must be zero:

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

### Step 7: Solve for \( \sin \alpha \)

This implies:

\[ 3 \sin \alpha = 0 \implies \sin \alpha = 0 \]

### Step 8: Find the values of \( \alpha \)

The solutions for \( \sin \alpha = 0 \) are:

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

### Final Answer

Thus, the real values of \( \alpha \) for which the expression is purely real are:

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

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