A value of \theta \, for which 2 + 3i\sin \theta \, \over 1 - 2i\,\,\sin \,\theta \, is purely imaginary, is :

\sin ^ - 1( 1 \over \sqrt 3 )\,

Complex Numbers PYQ — JEE Main 2016 · Shift 0 · Q22 with Solution

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Correct answer: B

Let $x=\sin\theta.$ Then the given complex number becomes $z=\frac{2+3ix}{1-2ix}.$ We want $z$ to be purely imaginary, so its real part must be $0$ .
Rationalize the denominator:

\frac{(2+3ix)(1+2ix)}{1+4x^2}.$$ Now expand the numerator:

(2+3ix)(1+2ix)=2+4ix+3ix+6i^2x^2.

Since $i^2=-1$,

(2+3ix)(1+2ix)=2+7ix-6x^2.

So,

z=\frac{(2-6x^2)+7ix}{1+4x^2}.

3. For $z$ to be purely imaginary, its real part must vanish:

\frac{2-6x^2}{1+4x^2}=0.

Since $1+4x^2>0$, we need $$2-6x^2=0.$$ Thus, $$6x^2=2 \Rightarrow x^2=\frac13.

Hence, $x=\pm \frac{1}{\sqrt3}.$ That is, $\sin\theta=\pm \frac{1}{\sqrt3}.$

A: $\sin^{-1}\left(\frac{\sqrt3}{4}\right)$ gives $\sin\theta=\frac{\sqrt3}{4}$ , not correct.
B: $\sin^{-1}\left(\frac{1}{\sqrt3}\right)$ gives $\sin\theta=\frac{1}{\sqrt3}$ , correct.
C: $\theta=\frac\pi3$ gives $\sin\theta=\frac{\sqrt3}{2}$ , not correct.
D: $\theta=\frac\pi6$ gives $\sin\theta=\frac12$ , not correct.

Therefore, the correct option is $\boxed{\text{B}}.$