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 \[z=\cos \theta +i\sin \cdot \theta |z|=1\] \[\because \] \[z=\frac{c+i}{c-i}\] \[\Rightarrow \]\[c=\frac{i(z+1)}{z-1}=\frac{i(\cos \theta +i\sin \theta +1)}{(\cos \theta +i\sin \theta -1)}\] \[=\frac{i\left( 2{{\cos }^{2}}\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2} \right)}{-2{{\sin }^{2}}\frac{\theta }{2}+2i\sin \frac{\theta }{2}\cos \frac{\theta }{2}}\] \[=\frac{i\left( \cos \frac{\theta }{2}+i\,\,\sin \frac{\theta }{2} \right)}{\left( -\sin \frac{\theta }{2}+i\cos \frac{\theta }{2} \right)}\cot \frac{\theta }{2}\] \[=\frac{i\left( \cos \frac{\theta }{2}+i\,\,\sin \frac{\theta }{2} \right)}{i\left( \cos \frac{\theta }{2}+i\sin \frac{\theta }{2} \right)}\cdot \frac{\cot \theta }{2}=\cot \frac{\theta }{2}\] \[c=\frac{1+\cos \theta }{\sin \theta }=\frac{1+a}{b}\]
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