#### Answer

find $\displaystyle \tan^{-1}\frac{1}{a}$ and add $\pi.$

#### Work Step by Step

$y=\tan^{-1} x$
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 $\cot y=x.$
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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 $\pi.$