Question

Write the value of tan⁻¹x + tan⁻¹$\left(\frac{1}{x}\right)$ for x > 0.

Answer 1

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right),xy<1$

$$\begin{gathered} \therefore {\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}={\tan }^{-1}\left(\frac{x+\frac{1}{x}}{1-x{\displaystyle \frac{1}{x}}}\right),x>0 \\ ={\tan }^{-1}\left(\frac{x^2+1}{{\displaystyle 0}}\right) \\ ={\tan }^{-1}\left(\infty \right) \\ ={\tan }^{-1}\left(\tan \frac{\pi }{2}\right) \\ =\frac{\pi }{2} \end{gathered}$$

$\therefore {\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=\frac{\pi }{2}$