The solution of \(\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}\) is

We have, \(\tan ^{-1} x+2 \cot ^{-1} x=\frac{2 \pi}{3}\)
\[
\Rightarrow \quad \tan ^{-1} x+2 \tan ^{-1} \frac{1}{x}=\frac{2 \pi}{3}
\]
\[
\Rightarrow \tan ^{-1} x+\tan ^{-1}\left(\frac{2\left(\frac{1}{x}\right)}{1-\left(\frac{1}{x}\right)^{2}}\right)=\frac{2 \pi}{3}
\]
\(\left[\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right]\)
\(\Rightarrow \quad \tan ^{-1} x+\tan ^{-1}\left(\frac{2 x}{x^{2}-1}\right)=\frac{2 \pi}{3}\)
\(\Rightarrow \quad \tan ^{-1}\left(\frac{x+\frac{2 x}{x^{2}-1}}{1-\frac{2 x^{2}}{x^{2}-1}}\right)=\frac{2 \pi}{3}\)
\(\Rightarrow \quad \frac{x^{3}-x+2 x}{x^{2}-1-2 x^{2}}=\tan \left(\frac{2 \pi}{3}\right)\)
\(\Rightarrow \quad \frac{x^{3}+x}{-1-x^{2}}=\tan \left(\frac{2 \pi}{3}\right)\)
\(\Rightarrow \quad \frac{x\left(x^{2}+1\right)}{-1\left(x^{2}+1\right)}=-\sqrt{3}\)
\(\Rightarrow \quad x=\sqrt{3}\)