Question

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

• A. $3$
• B. $\sqrt{3}$
• C. $\sqrt{2}$
• D. $\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Answer 1

Answer: (B)

$\sqrt{3}$

Given, ${\tan }^{-1}\left(x\right)+{\cot }^{-1}\left(x\right)+{\cot }^{-1}\left(x\right)=\frac{2\pi }{3}$
$\frac{\pi }{2}+{\cot }^{-1}\left(x\right)=\frac{2\pi }{3}$
${\cot }^{-1}\left(x\right)=\frac{\pi }{6}$
$x=\cot \left(\frac{\pi }{6}\right)$
$\therefore x=\sqrt{3}$