Browse Questions

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

Can you answer this question?

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 $x=cot\theta,\:\:\Rightarrow\:\theta=cot^{-1}x$
$cot^{-1}(-x)=cot^{-1}(-cot\theta)$
$=cot^{-1}cot(\pi-\theta)$
$=\pi-\theta=\pi-cot^{-1}x$

answered Feb 18, 2013
edited Mar 10, 2013