A) \[2\]
B) \[-1\]
C) \[1\]
D) \[-2\]
Correct Answer: C
Solution :
Given that, \[{{\tan }^{-1}}\,(1-x),\,\,\,{{\tan }^{-1}}x\] and \[{{\tan }^{-1}}\,(1+x)\] are in AP, then \[2{{\tan }^{-1}}x={{\tan }^{-1}}\,(1-x)+{{\tan }^{-1}}\,(1+x)\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \frac{1-x+1+x}{1-(1-x)\,(1+x)} \right)\] \[\Rightarrow \] \[{{\tan }^{-1}}\,\left( \frac{2x}{1-{{x}^{2}}} \right)={{\tan }^{-1}}\left( \frac{2}{{{x}^{2}}} \right)\] \[\Rightarrow \] \[\frac{2x}{1-{{x}^{2}}}=\frac{2}{{{x}^{2}}}\] \[\Rightarrow \] \[{{x}^{3}}=1-{{x}^{2}}\] \[\Rightarrow \] \[{{x}^{3}}+{{x}^{2}}=1\]
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