A) not defined
B) \[\sqrt{3}\]
C) \[\pm \,1\]
D) \[0\]
E) None of these
Correct Answer: E
Solution :
We have, \[{{\tan }^{-1}}\left( \frac{1}{x} \right)=\pi +{{\tan }^{-1}}x,\,0<x<1\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{1}{x} \right)-{{\tan }^{-1}}x=\pi \] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{\frac{1}{x}-x}{1+\frac{1}{x}.x} \right)=\pi \] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{1-{{x}^{2}}}{x(1+1)} \right)=\pi \] \[\Rightarrow \] \[\frac{1-{{x}^{2}}}{2x}=\tan \pi \] \[\Rightarrow \] \[\frac{1-{{x}^{2}}}{2x}=0\] \[\Rightarrow \] \[1-{{x}^{2}}=0\,\Rightarrow {{x}^{2}}=1\,\Rightarrow x=\pm 1\] but given, \[0<x<1\]
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