Question

If ${\tan }^{-1}x=\frac{\mathrm{\pi }}{4}-{\tan }^{-1}\frac{1}{3},$ then x = ______________________.

Answer 1

$$\begin{gathered} {\tan }^{-1}x=\frac{\mathrm{\pi }}{4}-{\tan }^{-1}\frac{1}{3} \\ \Rightarrow {\tan }^{-1}x={\tan }^{-1}1-{\tan }^{-1}\frac{1}{3} \\ \Rightarrow {\tan }^{-1}x={\tan }^{-1}\left(\frac{1-{\displaystyle \frac{1}{3}}}{1+1\times {\displaystyle \frac{1}{3}}}\right)\left[{\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right),xy>-1\right] \end{gathered}$$
$$\begin{gathered} \Rightarrow {\tan }^{-1}x={\tan }^{-1}\left(\frac{{\displaystyle \frac{2}{3}}}{{\displaystyle \frac{4}{3}}}\right) \\ \Rightarrow {\tan }^{-1}x={\tan }^{-1}\frac{1}{2} \\ \Rightarrow x=\frac{1}{2} \end{gathered}$$

If ${\tan }^{-1}x=\frac{\mathrm{\pi }}{4}-{\tan }^{-1}\frac{1}{3},$ then x = $\underline{\frac{1}{2}}$.