Answer
Using a calculator, we calculate
$y=\displaystyle \tan^{-1}\frac{1}{a} $and add $\pi$
Work Step by Step
(*) lnverse Tangent Function:
$y=\tan^{-1}x$ or $y=$ arctan $x$ means that $x=\tan y$, for $-\displaystyle \frac{\pi}{2} < y < \frac{\pi}{2}$.
(**) lnverse Cotangent Function:
$y=\cot^{-1}x$ or $y=$ arccot $x$ means that $x=\cot y$, for $ 0 < y < \pi$
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If a is negative,
$y=\cot^{-1}a$ means that $a=\cot y$, where $\displaystyle \frac{\pi}{2} < y < \pi$ (second quadrant)
Also, if a is negative
$y=\displaystyle \tan^{-1}\frac{1}{a}$ that $\displaystyle \frac{1}{a}=\tan y$, for $-\displaystyle \frac{\pi}{2} < y < 0$ (fourth quadrant)
Calculating $\displaystyle \tan^{-1}\frac{1}{a}$, we get a number from Q.IV.
We need to find a number from Q.II which will be in the domain of $\cot^{-1}$.
We know the period for tan and cot, $\tan y=\tan(y+\pi)$,
$-\displaystyle \frac{\pi}{2} < y < 0\qquad /+\pi$
$\displaystyle \frac{\pi}{2} < y+\pi < \pi$
So, we calculate $y=\displaystyle \tan^{-1}\frac{1}{a}, $which is in quadrant IV,
add $\pi$ to $y $so that
$ y+\pi$ is in quadrant $II, $and
$\cot(y+\pi)=a $, that is
$\displaystyle \tan^{-1}\frac{1}{a}+\pi=\cot^{-1}a$
