Question

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

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

Answer 1

The correct option is B $2$

Let $y={\text{tan}}^{-1}(\frac{1}{2+\sqrt{3}})⟹\tan y=\frac{1}{2+\sqrt{3}}$

$\tan 3y=\frac{3\tan y-{\tan }^{3}y}{1+3{\tan }^{2}y}=\frac{\frac{1}{2+\sqrt{3}}(3-\frac{1}{(2+\sqrt{3}{)}^2})}{1+\frac{3}{(2+\sqrt{3}{)}^2}}=\frac{\frac{1}{2+\sqrt{3}}(20+12\sqrt{3})}{4+4\sqrt{3}}=\frac{20+12\sqrt{3}}{20+12\sqrt{3}}=1$

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

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

$⟹x=2$