$\cot^{-1}(x)=\pi+\tan^{-1}(1/x)$ when $x<0$

Question

My book says $\cot^{-1}(x)=\pi+\tan^{-1}(1/x)$ when $x<0$ but when I made these two plots on wolfram alpha they look exactly same even when $x<0$. Why is this happening?

1) Plot of $\tan^{-1}(1/x)$

2) Plot of $\cot^{-1}(x)$

Can someone kindly explain this to me.Thanks.

See http://math.stackexchange.com/questions/304399/are-mathrmarccotx-and-arctan1-x-the-same-function and http://math.stackexchange.com/questions/610261/simplifying-an-arctan-equation?lq=1 — lab bhattacharjee, Apr 19 '15 at 18:20

abel · Answer 1 · 2015-04-19T15:26:42.377

1

it depends on the domain of the functions in question. it is customary to take the domain of $tan$ to be $(-\pi/2, \pi/2)$ and the domain of $\cot$ to be $(0, \pi).$

it is easier to see this on the unit circle. draw a line with slope $\frac 1m.$ it will cut the unit circle at two diametrically opposite points; one in the fourth quadrant, the other in the second quadrant. the signed arclength in the fourth quadrant correspond to $\tan^{-1}\left(\frac 1m\right).$ the one in the second quadrant correspond to $\cot^{-1}(m).$

suppose you take $m < 0,$ then you will find $-\pi/2 < t < 0$ such that $\tan(t) = \frac 1m, \cot(\pi + t) = m.$ that is $$\tan^{-1}\left(\frac 1m \right) = t, \pi + t = \cot^{-1}m \to \cot^{-1}m = \pi + tan^{-1}\left(\frac 1m\right), m < 0.$$

edited Apr 19 '15 at 15:26

answered Apr 19 '15 at 14:55

abel

29,170

The functions arctan and arccot are defined on the whole real line; you are talking about the codomain, perhaps. – egreg Apr 19 '15 at 15:09
Please explain a bit more clearly ! – Apr 19 '15 at 15:15
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@SanchayanDutta, i have edited my answer. please see if it helps see better. – abel Apr 19 '15 at 15:20
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@SanchayanDutta, you are welcome. – abel Apr 19 '15 at 15:27

score 0 · Accepted Answer · answered Apr 19 '15 at 15:16

The value of $\arctan x$ are (usually) in $(-\pi/2,\pi/2)$, while the values of $\def\arccot{\operatorname{arccot}}\arccot x$ are taken in $(0,\pi)$.

Now, since $\arctan(1/x)\in(-\pi/2,0)$, we have $\pi+\arctan(1/x)\in(\pi/2,\pi)$, which is the range of $\arccot x$ when $x<0$. Thus we have just to compute the cotangent of the right hand side term: $$ \cot\left(\pi+\arctan\frac{1}{x}\right)=\cot\arctan\frac{1}{x}= \frac{1}{\tan\arctan\dfrac{1}{x}}=\frac{1}{1/x}=x. $$ Since the cotangent of both sides is $x$, we are done.

$\cot^{-1}(x)=\pi+\tan^{-1}(1/x)$ when $x<0$

2 Answers2