Cfrgtkky

Right triangle with legs $1$ and $x$, just remembering the sum of all the angles of a triangle, is $pi$. In case of a right triangle, $alpha+beta={piover2}$ for any $x$.

Here's a proof
in the case when
$x > 0$.

Draw a right triangle
with legs $1$ and $x$,
with the leg of length $x$
opposite angle $A$.

Then
$tan(A) = x$
and
$tan(B) = 1/x$,
so $A = arctan(x)$
and
$B = arctan(1/x)$.

Since $A+B = pi/2$,
$arctan(x)+arctan(1/x)
= pi/2
$.

The tangent function should be defined to take the value $infty$ at $pmpi/2$, and this $infty$ is neither $+infty$ or $-infty$, but is the $infty$ that is approached by going either in the positive direction or the negative direction. That makes the tangent function everywhere continuous, including continuity at $pmpi/2$.

If one also identifies $+pi/2$ with $-pi/2$ so that the domain of the tangent function is topologically a circle one of whose points is $+pi/2=-pi/2$, then the tangent function is one-to-one. There is only one point in its domain that maps to $infty$, namely $+pi/2=-pi/2$.

After that, there remains the question of whether the standard identity for the tangent of a sum applies when $infty$ occurs among the values of the functions involved. To address that we should also take $infty$ (not $+infty$ and not $-infty$) to be the value of a rational function wherever it has a vertical asymptote. This makes rational functions everywhere continuous. Then we have
begin{align}
tan(alpha+beta) & overset{hugetext{?}}= frac{sin(alpha+beta)}{cos(alpha+beta)} tag 1 \[10pt]
& {} = frac{sinalphacosbeta+cosalphasinbeta}{cosalphacosbeta-sinalphasinbeta}. tag 2
end{align}
Notice that in $(1)$, the sine and cosine cannot both be $0$, so we need not consider what happens then.

In the standard argument, we divide both the numerator and the denominator by $cosalphacosbeta$, getting this:
$$
frac{sinalphacosbeta+cosalphasinbeta}{cosalphacosbeta-sinalphasinbeta} = frac{frac{sinalpha}{cosalpha} - frac{sinbeta}{cosbeta}}{1 - frac{sinalpha}{cosalpha}cdotfrac{sinbeta}{cosbeta}} tag 3
$$
All this is valid when there are no $0$s in denominators. We need to deal with $0$s in denominators.

First consider the case in which the denominator in $(1)$ is $0$ but $cosalphane0necosbeta$. Then the denominator on the right side of $(3)$ is $0$. But the numerator on the right is not $0$, since the numerator on the left is not $0$; the numerator on the left side of $(3)$ cannot be $0$ since $sin(alpha+beta) ne 0$. Hence the standard identity holds in that case.

Next consider the case where $cos(alpha+beta)=0$ and $cosalpha=0$. Then

$$
fracpi2 = - fracpi2 = alpha+beta = fracpi2+beta = -fracpi2 + beta
$$
and we're done.

You can't compute $tantheta$ when $theta=pmfracpi2$, but you can compute $costheta$ instead.
For $a,bin mathbb{R}$ with $abne 1$ we have
begin{eqnarray}
cos^2(arctan(a)+arctan(b))&=&frac{1}{1+tan^2(arctan(a)+arctan(b))}=frac{1}{1+left(frac{a+b}{1-ab}right)^2}\
&=&frac{(1-ab)^2}{(1-ab)^2+(ab)^2}.
end{eqnarray}
It follows that
$$
cos^2(f(x))=lim_{ato x,bto x^{-1}}cos^2(arctan(a)+arctan(b))=lim_{ato x,bto x^{-1}}frac{(1-ab)^2}{(1-ab)^2+(ab)^2}=0,
$$
i.e. $f(x)=pmfracpi2$. Since $f$ is continuous on $(-infty,0)$ and on $(0,infty)$, we deduce that
$$
f(x)=f(-1)=2arctan(-1)=-fracpi2 quad forall x<0,
$$
and
$$
f(x)=-f(-x)=fracpi2 quad forall x>0
$$

Added
The function $f: xmapsto f(x)=arctan(x)+arctan(x^{-1})$ defined and differentiable of $mathbb{R}setminus{0}$. For every $xne 0$ we have
$$
f'(x)=dfrac{1}{1+x^2}+dfrac{-x^{-2}}{1+x^{-2}}=dfrac{1}{1+x^2}-dfrac{1}{x^2+1}=0
$$
Therefore, $f$ is constant on each connected component of $mathbb{R}setminus{0}$.
Since
$$
f(1)=2arctan(1)=2dfrac{pi}{4}=dfrac{pi}{2},quad f(-1)=-f(1)=-dfrac{pi}{2},
$$
it follows that
$$
arctan(x)+arctanleft(dfrac{1}{x}right)=begin{cases}-dfrac{pi}{2} &text{ if } x<0\
dfrac{pi}{2} &text{ if } x>0
end{cases}.
$$

I just figured I'd throw this in.

Let $f(x) = arctan(x) + arctan(1/x)$ for all $x in (0, infty)$.

Then $f'(x) = dfrac{1}{1+x^2} - dfrac{dfrac{1}{x^2}}{1 + dfrac{1}{x^2}} = 0$.

Hence $f(x)$ is constant on $(0, infty)$.

Since $f(1) = dfrac{pi}{4} + dfrac{pi}{4} = dfrac{pi}{2}$,
we conclude that

$f(x) = dfrac{pi}{2}$ for all $x in (0, infty)$.

Addendum

If you're not ready for calculus, for the same
$x in (0, infty)$,
Consider the point
$P = (1, x)$, in the first quadrant, with corresponding angle
$0 lt theta lt dfrac{pi}{2}$.

Let $hat{theta} = dfrac{pi}{2} - theta$.
Then, also, $0 lt hat{theta} lt dfrac{pi}{2}$ and
$tan(hat theta)
= tan left( dfrac{pi}{2} - theta right)
= cot theta = dfrac 1x$

It follows that
$arctan x + arctan dfrac 1x
= theta + hat theta
= dfrac{pi}{2}$

For all $x in (-infty, 0)$, we have

$arctan x + arctan dfrac 1x =
-left(arctan(-x) + arctan left(-dfrac 1x right) right) =
-f(-x) = -dfrac{pi}{2}$.

Your logic is entirely correct.

The functions $tan^{-1} x $ and $tan^{-1} (1/x ) $ are both odd functions. Accordingly their additive $tan^{-1} x + tan^{-1} (1/x ) $ must be odd with $ + pi/2 $ value for positive arguments and $ - pi/2 $ value for negative arguments.

This function results by subtracting $ pi/2$ from the step function 0 to $pi$ jump at x = 0 and so making it an odd function of opposite sign for arguments of opposite sign, defining the $ f(x) $ to be $pi/2 $ for $ x>0, 0$ for $ x=0 $ and $ - pi/2 $ for $ x<0. $

An alternative proof using calculus.

Let $f(x) = arctan(x) + arctan(1/x)$ for all real $x$ except $0$.

$$frac {df(x)} {dx}= frac{1}{x^2+1} + frac {1}{1+1/x^2}*frac{-1}{x^2} = frac{1}{x^2+1} - frac{1}{x^2+1} = 0 $$

Integrating on both sides we get

$$f(x) = arctan(x) + arctan(1/x) = C,$$ where $C$ is a constant.

Since the function is discontinuous at $x=0$ the constant value can be different on both sides.

It can be obtained for $x>0$ by for example setting $x=1$:
$$f(1) = arctan(1) + arctan(1) = pi/4 + pi/4 = pi/2 $$
Similarly, for $x=-1$:
$$f(1) = arctan(-1) + arctan(-1) = -pi/4 -pi/4 = -pi/2 $$

So, in conclusion, $$arctan(x) + arctan(1/x) = pi/2 $$ for $x>0$
and
$$arctan(x) + arctan(1/x) = -pi/2 $$ for $x<0$.