Question

Find $x$ if ${\tan }^{-1}x+2{\cot }^{-1}x=\frac{2\pi }{3}$

Answer 1

$\to$ As $co{t}^{-1}x=ta{n}^{-1}\left(\frac{1}{x}\right)$

And $2ta{n}^{-1}\left(\frac{1}{x}\right)={\tan }^{-1}\left(\frac{\frac{2}{x}}{1-\frac{1}{x^2}}\right)$

$={\tan }^{-1}(\frac{2}{x}\times \frac{x^2}{x^2-1})={\tan }^{-1}\left(\frac{2x}{x^2-1}\right)$

${\tan }^{-1}x+{\tan }^{-1}\left(\frac{2x}{x^2-1}\right)={\tan }^{-1}\left(\frac{x+\frac{2x}{x^2-1}}{1-\frac{2x^2}{x^2-1}}\right)$

$={\tan }^{-1}\left(\frac{x^3-x+2x}{x^2-1-2x^2}\right)={\tan }^{-1}\left(\frac{x^3+x}{-x^2-1}\right)$

$=-{\tan }^{-1}\left(\frac{x(x^2+1)}{x^2+1}\right)=-{\tan }^{-1}x$

$\Rightarrow ta{n}^{-1}x=\frac{-2\pi }{3}$

$\Rightarrow x=\tan \left(\frac{-2\pi }{3}\right)=-\tan (\pi -\frac{\pi }{3})$

$x=\frac{1}{\sqrt{3}}$