A) \[2n\pi \]
B) \[n\pi +\frac{\pi }{2}\]
C) \[n\pi \]
D) None of these [Where \[n\] is an integer]
Correct Answer: C
Solution :
\[\frac{(3+2i\sin \theta )(1+2i\sin \theta )}{(1-2i\sin \theta )(1+2i\sin \theta )}\]= \[\left( \frac{3-4{{\sin }^{2}}\theta }{1+4{{\sin }^{2}}\theta } \right)+i\left( \frac{8\sin \theta }{1+4{{\sin }^{2}}\theta } \right)\] 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\[\theta \].
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