A value of $$\theta$$ for which $$\frac{2 + 3i\sin\theta}{1 - 2i\sin\theta}$$ is purely imaginary, is

We want to find those real numbers $$\theta$$ for which the complex quantity

$$\dfrac{2 + 3i\sin\theta}{1 - 2i\sin\theta}$$

is purely imaginary. A complex number is purely imaginary when its real part is zero. Hence our task is to extract the real part of the given fraction and set it equal to $$0$$.

First of all, having a complex number in the denominator is awkward, so we rationalise the denominator. The standard method is to multiply the numerator and the denominator by the complex conjugate of the denominator. Here the denominator is $$1 - 2i\sin\theta$$, and its conjugate is $$1 + 2i\sin\theta$$. Therefore we write

$$ \dfrac{2 + 3i\sin\theta}{1 - 2i\sin\theta} \;=\; \dfrac{\,\bigl(2 + 3i\sin\theta\bigr)\bigl(1 + 2i\sin\theta\bigr)\,} {\,\bigl(1 - 2i\sin\theta\bigr)\bigl(1 + 2i\sin\theta\bigr)\,}. $$

Now we simplify the denominator using the identity $$\bigl(a - ib\bigr)\bigl(a + ib\bigr)=a^{2}+b^{2}$$:

$$ \bigl(1 - 2i\sin\theta\bigr)\bigl(1 + 2i\sin\theta\bigr) = 1^{2} + \bigl(2\sin\theta\bigr)^{2} = 1 + 4\sin^{2}\theta. $$

Next we expand the numerator. We treat $$\sin\theta$$ as a single symbol to keep the algebra neat. Denote $$\sin\theta = s$$ for brevity. With this notation the numerator is

$$ \bigl(2 + 3is\bigr)\bigl(1 + 2is\bigr). $$

We multiply term by term:

$$ \begin{aligned} (2)(1) & = 2,\\[2pt] (2)(2is) & = 4is,\\[2pt] (3is)(1) & = 3is,\\[2pt] (3is)(2is) & = 6i^{2}s^{2}. \end{aligned} $$

Adding these four pieces gives

$$ 2 + 4is + 3is + 6i^{2}s^{2}. $$

Because $$i^{2}=-1$$, the last term becomes $$6(-1)s^{2}=-6s^{2}$$. Collecting real and imaginary parts we obtain

$$ \bigl(2 - 6s^{2}\bigr) + 7is. $$

So, after rationalising, the entire complex number is

$$ \dfrac{\bigl(2 - 6s^{2}\bigr) + 7is}{1 + 4s^{2}}. $$

Writing it explicitly as “real part + imaginary part” we get

$$ \dfrac{2 - 6s^{2}}{1 + 4s^{2}} \;+\; i\,\dfrac{7s}{1 + 4s^{2}}. $$

The fraction will be purely imaginary exactly when its real part equals zero. Therefore we must solve

$$ \dfrac{2 - 6s^{2}}{1 + 4s^{2}} = 0. $$

The denominator $$1 + 4s^{2}$$ is always positive, so the only way for the quotient to be zero is for the numerator to be zero:

$$ 2 - 6s^{2} = 0. $$

Rearranging, we have

$$ 6s^{2} = 2 \;\Longrightarrow\; s^{2} = \dfrac{2}{6} = \dfrac{1}{3}. $$

Taking square roots we obtain

$$ s = \pm\dfrac{1}{\sqrt{3}}. $$

Recall that $$s = \sin\theta$$, so the required condition is

$$ \sin\theta = \pm\dfrac{1}{\sqrt{3}}. $$

The options list only the positive value $$\sin^{-1}\!\left(\dfrac{1}{\sqrt{3}}\right)$$, and that is enough because the question asks for “a value” of $$\theta$$. Thus we select the choice that matches $$\sin^{-1}\!\bigl(1/\sqrt{3}\bigr)$$.

Hence, the correct answer is Option B.