A value of for which `(2+3isintheta)/(1-2isintheta)` purely imaginary, is : (1) `pi/3` (2) `pi/6` (3) `sin^(-1)((sqrt(3))/4)` (4) `sin^(-1)(1/(sqrt(3)))`

Options

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

Explanation

To solve the problem of finding 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: Write down the expression
We start with the expression:
\[ z = \frac{2 + 3i \sin \theta}{1 - 2i \sin \theta} \]

### Step 2: Rationalize the denominator
To eliminate the imaginary part in the denominator, we multiply the numerator and 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: Calculate the denominator
The denominator simplifies as follows:
\[ (1 - 2i \sin \theta)(1 + 2i \sin \theta) = 1^2 - (2i \sin \theta)^2 = 1 + 4 \sin^2 \theta \]

### Step 4: Calculate the numerator
Now, we calculate the numerator:
\[ (2 + 3i \sin \theta)(1 + 2i \sin \theta) = 2 \cdot 1 + 2 \cdot 2i \sin \theta + 3i \sin \theta \cdot 1 + 3i \sin \theta \cdot 2i \sin \theta \]
This expands to:
\[ 2 + 4i \sin \theta + 3i \sin \theta - 6 \sin^2 \theta = 2 - 6 \sin^2 \theta + (4i \sin \theta + 3i \sin \theta) \]
Combining the imaginary parts gives:
\[ 2 - 6 \sin^2 \theta + 7i \sin \theta \]

### Step 5: Combine the results
Now we can write \( z \) as:
\[ z = \frac{2 - 6 \sin^2 \theta + 7i \sin \theta}{1 + 4 \sin^2 \theta} \]

### Step 6: Separate real and imaginary parts
The real part of \( z \) is:
\[ \text{Re}(z) = \frac{2 - 6 \sin^2 \theta}{1 + 4 \sin^2 \theta} \]
The imaginary part of \( z \) is:
\[ \text{Im}(z) = \frac{7 \sin \theta}{1 + 4 \sin^2 \theta} \]

### Step 7: Set the real part to zero
For \( z \) to be purely imaginary, the real part must be zero:
\[ 2 - 6 \sin^2 \theta = 0 \]

### Step 8: Solve for \( \sin \theta \)
Rearranging gives:
\[ 6 \sin^2 \theta = 2 \quad \Rightarrow \quad \sin^2 \theta = \frac{1}{3} \quad \Rightarrow \quad \sin \theta = \frac{1}{\sqrt{3}} \]

### Step 9: Find \( \theta \)
Now, we find \( \theta \):
\[ \theta = \sin^{-1} \left( \frac{1}{\sqrt{3}} \right) \]

### Conclusion
Thus, the value of \( \theta \) for which the expression is purely imaginary is:
\[ \theta = \sin^{-1} \left( \frac{1}{\sqrt{3}} \right) \]
The correct option is (4) \( \sin^{-1} \left( \frac{1}{\sqrt{3}} \right) \).