Question

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

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

Answer 1

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

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

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

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

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

G.E is purely imaginary

Hence real part$=0$

$2-6{\sin }^2\theta =0$

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

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

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

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