A value of `theta` for which `(2+3isintheta)/(1-2isintheta)` is purely imaginary, is

Options

A. `pi/6`
B. `sin^(-1)(sqrt(3)/4)`
C. `sin^(-1)((1)/sqrt(3))`
D. `pi/3`

Answer

To find the value of \( \theta \) for which the expression \( \frac{2 + 3i \sin \theta}{1 - 2i \sin \theta} \) is purely imaginary, we will follow these steps:

### Step 1: Define the complex number
Let
\[ z = \frac{2 + 3i \sin \theta}{1 - 2i \sin \theta} \]
We need to find \( \theta \) such that \( z \) is purely imaginary. This means that the real part of \( z \) must be zero.

### Step 2: Multiply by the conjugate
To simplify the expression, multiply the numerator and the denominator by the conjugate of the denominator:
\[ z = \frac{(2 + 3i \sin \theta)(1 + 2i \sin \theta)}{(1 - 2i \sin \theta)(1 + 2i \sin \theta)} \]

### Step 3: Simplify the denominator
The denominator simplifies as follows:
\[ (1 - 2i \sin \theta)(1 + 2i \sin \theta) = 1^2 - (2i \sin \theta)^2 = 1 + 4 \sin^2 \theta \]

### Step 4: Simplify the numerator
Now simplify the numerator:
\[ (2 + 3i \sin \theta)(1 + 2i \sin \theta) = 2 \cdot 1 + 2 \cdot 2i \sin \theta + 3i \sin \theta \cdot 1 + 3i \sin \theta \cdot 2i \sin \theta \]
\[ = 2 + 4i \sin \theta + 3i \sin \theta - 6 \sin^2 \theta \]
\[ = (2 - 6 \sin^2 \theta) + (4i \sin \theta + 3i \sin \theta) = (2 - 6 \sin^2 \theta) + 7i \sin \theta \]

### Step 5: Combine the results
Putting it all together, we have:
\[ z = \frac{(2 - 6 \sin^2 \theta) + 7i \sin \theta}{1 + 4 \sin^2 \theta} \]

### Step 6: Set the real part to zero
For \( z \) to be purely imaginary, the real part must be zero:
\[ \frac{2 - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta} = 0 \]
This implies:
\[ 2 - 6 \sin^2 \theta = 0 \]

### Step 7: Solve for \( \sin^2 \theta \)
Rearranging gives:
\[ 6 \sin^2 \theta = 2 \implies \sin^2 \theta = \frac{1}{3} \]

### Step 8: Find \( \sin \theta \)
Taking the square root:
\[ \sin \theta = \frac{1}{\sqrt{3}} \]

### Step 9: Find \( \theta \)
The values of \( \theta \) that satisfy this equation are:
\[ \theta = \sin^{-1}\left(\frac{1}{\sqrt{3}}\right) \]
This corresponds to:
\[ \theta = \frac{\pi}{6} \quad \text{and} \quad \theta = \frac{5\pi}{6} \quad \text{(in the range [0, 2π])} \]

### Final Answer
The values of \( \theta \) for which \( \frac{2 + 3i \sin \theta}{1 - 2i \sin \theta} \) is purely imaginary are:
\[ \theta = \frac{\pi}{6}, \frac{5\pi}{6} \]