A) \[\frac{\pi }{2}-\frac{\theta }{2}\]
B) \[\frac{\pi }{2}+\frac{\theta }{2}\]
C) \[\frac{\pi }{4}-\frac{\theta }{2}\]
D) \[\frac{\pi }{4}+\frac{\theta }{2}\]
Correct Answer: D
Solution :
Given complex number is \[\left( 1- sin\,\theta \right) +\,\,i cos\,\theta = a + ib\] \[\operatorname{Argument} = tan \theta = \frac{b}{a}\] \[\Rightarrow \,\,\,\tan \theta =\frac{\cos \theta }{1-\sin \theta }\] \[=\,\,\frac{{{\cos }^{2}}\frac{\theta }{2}-{{\sin }^{2}}\frac{\theta }{2}}{{{\sin }^{2}}\frac{\theta }{2}+{{\cos }^{2}}\frac{\theta }{2}-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\] \[=\,\,\,\frac{\left( \cos \frac{\theta }{2}-\sin \frac{\theta }{2} \right)\left( \cos \frac{\theta }{2}+\sin \frac{\theta }{2} \right)}{{{\left( \sin \,\frac{\theta }{2}-\cos \frac{\theta }{2} \right)}^{2}}}\] \[=\,\,\,\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}\] \[=\,\,\,\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}=\frac{\tan \frac{\pi }{4}+\tan \frac{\theta }{2}}{1-\tan \frac{\pi }{4}\,\tan \frac{\theta }{2}}\] \[\tan \theta =\tan \left( \frac{\pi }{4}+\frac{\theta }{2} \right)\] Hence, \[\operatorname{argument}=\frac{\pi }{4}+\frac{\theta }{2}\]
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