A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) \[\sqrt{2}\]
D) \[\frac{1}{\sqrt{2}}\]
Correct Answer: B
Solution :
Let \[z=\frac{1}{1+\cos \theta +i\sin \theta }\] \[=\frac{1}{2{{\cos }^{2}}\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\] \[=\frac{1}{2\cos \frac{\theta }{2}\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}\] \[=\frac{\cos \left( -\frac{\theta }{2} \right)+i\sin \left( -\frac{\theta }{2} \right)}{2\cos \frac{\theta }{2}}=\frac{\cos \frac{\theta }{2}-i\sin \frac{\theta }{2}}{2\cos \frac{\theta }{2}}\] [using De-Moivres Theorem] \[=\frac{1}{2}\,\left[ 1-i\tan \frac{\theta }{2} \right]\] Real of \[z=\frac{1}{2}\] real part of \[\left[ 1-i\tan \frac{\theta }{2} \right]\] \[=\frac{1}{2}\,(1)=\frac{1}{2}\]
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