$\cot^{-1}(x)=\pi+\tan^{-1}(1/x)$ when $x<0$ 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.
 A: 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 $$\tan^{-1}\left(\frac 1m \right) = t, \pi + t = \cot^{-1}m \to \cot^{-1}m = \pi + tan^{-1}\left(\frac 1m\right), m < 0.$$
A: 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.
