• # question_answer $\frac{3+2i\sin \theta }{1-2i\sin \theta }$will be real, if $\theta$ =    [IIT 1976; EAMCET 2002] A) $2n\pi$ B) $n\pi +\frac{\pi }{2}$ C) $n\pi$ D) None of these [Where $n$ is an integer]

$\frac{(3+2i\sin \theta )(1+2i\sin \theta )}{(1-2i\sin \theta )(1+2i\sin \theta )}$= $\left( \frac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta } \right)+i\left( \frac{8\sin \theta }{1+4{{\sin }^{2}}\theta } \right)$ Now, since it is real, therefore Im $(z)=0$ Þ $\frac{8\sin \theta }{1+4{{\sin }^{2}}\theta }$= 0 Þ $\sin \theta =0$, $\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$\theta$.