# The value of $\cot^{-1}(-x)$ for all $x\;\in\;R$ in terms of $\cot^{-1}x$ is ____________.

Toolbox:
• Principal interval of cot is $$(0,\pi)$$
• $$cot(\pi-\theta)=-cot\theta$$
$$Ans:\:\: \pi - cot^{-1}x$$
Since cot is -ve in the interval $$(\frac{\pi}{2},\pi)$$, the angle is to be reduced and brought within this interval.
Let $$x=cot\theta,\:\:\Rightarrow\:\theta=cot^{-1}x$$
$$cot^{-1}(-x)=cot^{-1}(-cot\theta)$$
$$=cot^{-1}cot(\pi-\theta)$$
$$=\pi-\theta=\pi-cot^{-1}x$$

edited Mar 10, 2013