JEE MAIN - Mathematics (2016 (Offline) - No. 3)
A value of $$\theta \,$$ for which $${{2 + 3i\sin \theta \,} \over {1 - 2i\,\,\sin \,\theta \,}}$$ is purely imaginary, is :
$${\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 4}} \right)$$
$${\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)\,$$
$${\pi \over 3}$$
$${\pi \over 6}$$
Explanation
Rationalizing the given expression
$${{\left( {2 + 3i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
For the given expression to be purely imaginary, real part of the above expression should be equal to zero.
$$ \Rightarrow {{2 - 6{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }} = 0$$
$$ \Rightarrow {\sin ^2}\theta = {1 \over 3}$$
$$ \Rightarrow \sin \theta = \pm {1 \over {\sqrt 3 }}$$
$${{\left( {2 + 3i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$$
For the given expression to be purely imaginary, real part of the above expression should be equal to zero.
$$ \Rightarrow {{2 - 6{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }} = 0$$
$$ \Rightarrow {\sin ^2}\theta = {1 \over 3}$$
$$ \Rightarrow \sin \theta = \pm {1 \over {\sqrt 3 }}$$
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