Question

The solution of $Ta{n}^{-1}x+2co{t}^{-1}x=\frac{2\pi }{3}$ is

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\sqrt{3}$
• D. $-\sqrt{3}$

Answer 1

Answer: (C)

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