A) \[\frac{1}{\cos \alpha }(\cos \alpha +i\sin \alpha )\]
B) \[\frac{1}{-\cos \alpha }[\cos \,(\pi +\alpha )+i\sin (\pi +\alpha )\]
C) \[\frac{1}{\cos \alpha }[\cos \,(2\pi +\alpha )+i\sin (2\pi +\alpha )]\]
D) None of these
Correct Answer: B
Solution :
\[\operatorname{z} = 1 + i tan \alpha = r \left( cos\,\theta +i sin\theta \right)\] \[\Rightarrow \,\,\,\operatorname{r} cos \theta = 1, r sin \theta = tan a \Rightarrow {{r}^{2}} = se{{c}^{2}}\alpha \] \[\Rightarrow \,\,\,r=\left| \sec \,\,\alpha \right|\,\,=\,\,\frac{1}{\left| \cos \,\,\alpha \right|}\] Since, \[- \pi < \alpha <-\frac{\pi }{2}\] \[\Rightarrow \,\,\,\cos \alpha < 0 \,\Rightarrow \,\, \left| cos \alpha \right| =-\,\,cos\,\alpha \] \[\therefore \,\,\,r=\frac{1}{-\cos \alpha }.\] Further, we get \[\cos \theta = - cos \alpha = cos \left( \pi + \alpha \right)\] Now, \[-\pi <\alpha <-\frac{\pi }{2}\Rightarrow \pi -\pi <\pi +\alpha <\pi -\frac{\pi }{2}\] \[\Rightarrow \,\, 0<\pi +\alpha <\frac{\pi }{2}\] [Converted to principal value] \[\therefore \,cos \theta = cos \left( \pi + a \right) \Rightarrow \theta =\pi +\alpha \] Hence \[\operatorname{z} = \frac{1}{-\cos \,\alpha }\,\left[ cos \left( \pi + \alpha \right) + i sin\,\left( \pi + \alpha \right) \right]\]
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