If \[{{\tan }^{-1}}x+2{{\cot }^{-1}}x=\frac{2\pi }{3},\]then x = [Karnataka CET 1999]

\[\frac{\sqrt{3}-1}{\sqrt{3}+1}\]

If \[{{\tan }^{-1}}x+2{{\cot }^{-1}}x=\frac{2\pi }{3},\]then x = [Karn

1.	If \[{{\tan }^{-1}}x+2{{\cot }^{-1}}x=\frac{2\pi }{3},\]then x = [Karnataka CET 1999]
A.	\[\sqrt{2}\]
B.	3
C.	\[\sqrt{3}\]
D.	\[\frac{\sqrt{3}-1}{\sqrt{3}+1}\]
Answer» D. \[\frac{\sqrt{3}-1}{\sqrt{3}+1}\]

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If \[{{\tan }^{-1}}x+2{{\cot }^{-1}}x=\frac{2\pi }{3},\]then x = [Karnataka CET 1999]

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