A) \[\frac{a}{b}\]
B) \[\frac{a-1}{b}\]
C) \[\frac{a+1}{b}\]
D) \[\frac{a+1}{b+1}\]
Correct Answer: C
Solution :
Let \[a=\,\cos \,\theta \] and \[b=\sin \,\theta \] \[\therefore \] \[z=\cos \,\theta +i\,\sin \,\theta \] \[(\because \,\,|\mathbf{z}|\,=\mathbf{1})\] \[\because \] \[z=\frac{c+i}{c-i}\] \[\therefore \] \[c=\frac{i(z+1)}{z-1}\,=\frac{i(\cos \,\theta +i\,\sin \,\theta +1)}{(\cos \theta +i\,\sin \,\theta -1)}\] \[=\frac{i\,\left\{ 2\,{{\cos }^{2}}\frac{\theta }{2}+2\,i\,\sin \,\frac{\theta }{2}\cdot \,\cos \,\frac{\theta }{2} \right\}}{\left\{ -2\,{{\sin }^{2}}\frac{\theta }{2}+2i\,\sin \,\frac{\theta }{2}\cdot \cos \,\frac{\theta }{2} \right\}}\] \[=i\,\frac{\cos \,\frac{\theta }{2}+\sin \,\frac{\theta }{2}}{\left( -\,\sin \,\frac{\theta }{2}+i\,\cos \,\frac{\theta }{2} \right)}\,\cdot \cot \,\frac{\theta }{2}\] \[=\frac{i}{i}\cdot \,\frac{\left( \cos \,\frac{\theta }{2}\,+i\,\sin \,\frac{\theta }{2} \right)}{\left( \cos \,\frac{\theta }{2}+\,i\,\sin \,\frac{\theta }{2} \right)}\cdot \,\cot \,\frac{\theta }{2}\] \[(\because \,\,{{i}^{2}}=1)\] \[=\cot \,\frac{\theta }{2}\] \[\Rightarrow \] \[c=\cot \,\frac{\theta }{2}\,=\frac{1+\cos \,\theta }{\sin \,\theta }=\frac{1+a}{b}\]
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