Question

If tan^-1x - cot^-1x = tan^-1$\sqrt{3}$, then x = _______________________.

Answer 1

$$\begin{gathered} {\tan }^{-1}x-{\cot }^{-1}x={\tan }^{-1}\sqrt{3} \\ \Rightarrow {\tan }^{-1}x-\left(\frac{\mathrm{\pi }}{2}-{\tan }^{-1}x\right)=\frac{\mathrm{\pi }}{3}\left({\tan }^{-1}x+{\cot }^{-1}x=\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow 2{\tan }^{-1}x=\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{3} \\ \Rightarrow {\tan }^{-1}x=\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6} \end{gathered}$$
$$\begin{gathered} \Rightarrow x=\tan \left(\frac{\mathrm{\pi }}{4}+\frac{\mathrm{\pi }}{6}\right) \\ \Rightarrow x=\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan {\displaystyle \frac{\mathrm{\pi }}{6}}}{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\times \tan {\displaystyle \frac{\mathrm{\pi }}{6}}} \\ \Rightarrow x=\frac{1+{\displaystyle \frac{1}{\sqrt{3}}}}{1-{\displaystyle \frac{1}{\sqrt{3}}}}=\frac{\sqrt{3}+1}{\sqrt{3}-1} \end{gathered}$$
$$\begin{gathered} \Rightarrow x=\frac{{\left(\sqrt{3}+1\right)}^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} \\ \Rightarrow x=\frac{4+2\sqrt{3}}{2} \\ \Rightarrow x=2+\sqrt{3} \end{gathered}$$

If tan⁻¹x − cot⁻¹x = tan⁻¹$\sqrt{3}$, then x = $\underline{2+\sqrt{3}}$.