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$\frac{{3 + 2i\sin \theta }}{{1 - 2i\sin \theta }}$will be real, if $\theta $ =
[Where $n$ is an integer]
$2n\pi $
$n\pi + \frac{\pi }{2}$
$n\pi $
None of these
Solution
(c) $\frac{{(3 + 2i\sin \theta )(1 + 2i\sin \theta )}}{{(1 – 2i\sin \theta )(1 + 2i\sin \theta )}}$= $\left( {\frac{{3 – 4{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}\theta }}} \right) + i\left( {\frac{{8\sin \theta }}{{1 + 4{{\sin }^2}\theta }}} \right)$
Now, since it is real, therefore Im $(z) = 0$
==> $\frac{{8\sin \theta }}{{1 + 4{{\sin }^2}\theta }}= 0$ ==> $\sin \theta = 0$, $\therefore $$\theta = n\pi $
where $n = 0$, $1, 2, 3, ……$
Trick : Check for $(a)$, if $n = 0,\theta = 0$ the given number is absolutely real but $(c)$ also satisfies this condition and in $(a)$ and $(c), (c)$ is most general value of $\theta $.
