Question

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

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. ${\sin }^{-1}\left(\frac{\sqrt{3}}{4}\right)$
• D. ${\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Answer 1

The correct option is D ${\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

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

Purely imaginary means real part $=0$

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

$=\frac{2+4i\sin \theta +3i\sin \theta -6{\sin }^2\theta }{1-(2i\sin \theta {)}^2}$

$=\frac{2+7i\sin \theta -6{\sin }^2\theta }{1+4{\sin }^2\theta }$

Real part is $=\frac{2-6{\sin }^2\theta }{1+4{\sin }^2\theta }$

This is equal to zero.

$\Rightarrow \frac{2-6{\sin }^2\theta }{1+4{\sin }^2\theta }=0$

So, $2=6{\sin }^2\theta$

${\sin }^2\theta =\frac{1}{3}$

$\sin \theta =\pm \frac{1}{\sqrt{3}}$

$\therefore \theta =si{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$