## Precalculus (6th Edition)

find $\displaystyle \tan^{-1}\frac{1}{a}$ and add $\pi.$
$y=\tan^{-1} x$ Domain: $(-\infty, \infty)$, Range:$(-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ y is the number from $(-\displaystyle \frac{\pi}{2}, \displaystyle \frac{\pi}{2})$ for which $\tan y=x$ $y=\cot^{-1}x$ Domain:$(-\infty, \infty)$, Range:$(0, \pi)$ y is the number from $(0, \pi)$ for which $\cot y=x.$ -------------------- So, if a is positive, then $y=\displaystyle \tan^{-1} \frac{1}{a}$is from $(0, \displaystyle \frac{\pi}{2}),$ $\displaystyle \tan y=\frac{1}{a}$, so $a=\cot y$, and $y$ belongs to the domain of $\cot^{-1}.$ If a is negative, $y=\displaystyle \tan^{-1} \frac{1}{a}$is from $(-\displaystyle \frac{\pi}{2}, 0)$, $\displaystyle \tan y=\frac{1}{a}$, but $y$ is not from the domain of $\cot^{-1}$. But since $\tan(y+\pi)$ also equals $\displaystyle \frac{1}{a},$ and $y+\pi$ IS in the domain of $\cot^{-1}$, $\cot(y+\pi)=a$, we find $\displaystyle \tan^{-1}\frac{1}{a}$ and add $\pi.$