Question

If $3{\tan }^{-1}\left(\frac{1}{2+\sqrt{3}}\right)-{\tan }^{-1}\left(\frac{1}{x}\right)={\tan }^{-1}\left(\frac{1}{3}\right)$, then x is equal to

• A. $1$
• B. $3$
• C. $2$
• D. $4$

Answer 1

The correct option is C $2$

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

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

or, ${\tan }^{-1}\left(\frac{12\sqrt{3}+20}{(2+\sqrt{3})(4+4\sqrt{3})}\right)-{\tan }^{-1}\left(\frac{1}{x}\right)={\tan }^{-1}\left(\frac{1}{3}\right)$

or, ${\tan }^{-1}\left(\frac{3\sqrt{3}+5}{(2+\sqrt{3})(1+\sqrt{3})}\right)-{\tan }^{-1}\left(\frac{1}{x}\right)={\tan }^{-1}\left(\frac{1}{3}\right)$

or, ${\tan }^{-1}\left(\frac{3\sqrt{3}+5}{3\sqrt{3}+5}\right)-{\tan }^{-1}\left(\frac{1}{x}\right)={\tan }^{-1}\left(\frac{1}{3}\right)$

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

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

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

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

$\Rightarrow x=2$.