A) \[-\frac{1}{\sqrt{3}}\]
B) \[\frac{1}{\sqrt{3}}\]
C) \[-\sqrt{3}\]
D) \[\sqrt{3}\]
Correct Answer: D
Solution :
Given,\[{{\tan }^{-1}}x+2{{\cot }^{-1}}x=\frac{2\pi }{3}\] \[\Rightarrow \] \[{{\tan }^{-1}}x+2{{\tan }^{-1}}\frac{1}{x}=\frac{2\pi }{3}\] \[\Rightarrow \] \[{{\tan }^{-1}}x+{{\tan }^{-1}}\left( \frac{2\left( \frac{1}{x} \right)}{1-{{\left( \frac{1}{x} \right)}^{2}}} \right)=\frac{2\pi }{3}\] \[\left[ \because \,\,2{{\tan }^{-1}}x={{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}} \right]\] \[\Rightarrow \] \[{{\tan }^{-1}}x+{{\tan }^{-1}}\left( \frac{2x}{{{x}^{2}}-1} \right)=\frac{2\pi }{3}\] \[\Rightarrow \] \[{{\tan }^{-1}}\left( \frac{x+\frac{2x}{{{x}^{2}}-1}}{1-\frac{2{{x}^{2}}}{{{x}^{2}}-1}} \right)=\frac{2\pi }{3}\] \[\Rightarrow \] \[\frac{{{x}^{3}}-x+2x}{{{x}^{2}}-1-2{{x}^{2}}}=\tan \left( \frac{2\pi }{3} \right)\] \[\Rightarrow \] \[\frac{{{x}^{3}}+x}{-1-{{x}^{2}}}=\tan \left( \frac{2\pi }{3} \right)\] \[\Rightarrow \] \[\frac{x({{x}^{2}}+1)}{-1({{x}^{2}}+1)}=-\sqrt{3}\] \[\Rightarrow \] \[x=\sqrt{3}\,\]
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