A) \[(2n+1)\frac{\pi }{2}\]
B) \[(n+1)\frac{\pi }{2}\]
C) \[n\pi \]
D) none of these
Correct Answer: C
Solution :
Key Idea: If \[z\]is purely real, then the coefficient of imaginary part will be zero. Let \[z=\frac{1-i\sin \alpha }{1+2i\,\sin \alpha }\] \[=\frac{1-i\sin \alpha }{1+2i\sin \alpha }\times \frac{(1-2i\,\sin \alpha )}{(1-2i\sin \alpha )}\] \[=\frac{1-2{{\sin }^{2}}\alpha -3i\sin \alpha }{1+4{{\sin }^{2}}\alpha }\] \[=\frac{(1-2{{\sin }^{2}}\alpha )-3i\,\sin \alpha }{1+4{{\sin }^{2}}\alpha }\] Since, z is real, therefore the coefficient of imaginary part will be zero. \[\Rightarrow \] \[3\sin \alpha =0\] \[\Rightarrow \] \[\alpha =n\pi ,\]where n is integer
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