Question

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

Answer 1

${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$
When $x<0,\frac{1}{x}<0$, then both are negative.
Let x = $-$ y, y>0
Then,
$$\begin{gathered} {\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}={\tan }^{-1}\left(-y\right)+{\tan }^{-1}\left(-\frac{1}{y}\right) \\ =-\left({\tan }^{-1}y+{\tan }^{-1}\frac{1}{y}\right) \\ =-{\tan }^{-1}\left(\frac{y+\frac{1}{y}}{1-y{\displaystyle \frac{1}{y}}}\right),y>0 \\ =-{\tan }^{-1}\left(\frac{y^2+1}{{\displaystyle 0}}\right) \\ =-{\tan }^{-1}\left(\infty \right) \\ =-{\tan }^{-1}\left(\tan \frac{\pi }{2}\right) \\ =-\frac{\pi }{2} \end{gathered}$$
$\therefore {\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}=-\frac{\pi }{2},x<0$