A) \[\cos \,x+i\,\sin \,x\]
B) \[m/2\]
C) 1
D) \[(m+1)/2\]
Correct Answer: C
Solution :
Let \[{{\cot }^{-1}}x=\theta \] Now, \[\left( \frac{xi+1}{xi-1} \right)=\,\frac{i\,\cot \,\theta +1}{i\,\cot \,\theta -1}=\frac{\cot \,\theta -i}{\cot \,\theta +i}\] \[=\frac{\cos \,\theta -i\,\sin \,\theta }{\cos \,\theta +i\,\sin \,\theta }=\frac{{{e}^{-i\theta }}}{{{e}^{i\theta }}}=\frac{1}{{{2}^{2/\theta }}}\] \[\Rightarrow \] \[{{e}^{2/\theta }}\left( \frac{xi+1}{xi-1} \right)=1\] \[\Rightarrow \] \[{{e}^{2mi\theta }}{{\left( \frac{xi+1}{xi-1} \right)}^{m}}=1\] \[\therefore \] \[{{e}^{2mi\,{{\cot }^{-1}}x}}{{\left( \frac{xi+1}{xi-1} \right)}^{m}}=1\]
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