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I know that two questions have already been asked about this exercise, but what I'm asking here is if this solution, which sounds rather strange to me, could possibly be correct. The problems is as follows:

Prove that $\,f(x)=\arctan(x)+\arctan(1/x)= \pi/2\,$ if $\,x>0\,$ and $\,-\pi/2\,$ if $\,x<0$.

What I did was this: first of all we note that, since $-\pi/2<\arctan y<\pi/2$ for all $y$, $-\pi<f(x)<\pi$. Now , consider $$ \tan\big(f(x)\big) = \tan\left(\arctan(x)+\arctan\left(\frac{1}{x} \right)\right) = \frac{\tan\left(\arctan(x)\right) + \tan\left(\arctan\left(\frac{1}{x} \right)\right)} {1 - \tan\big(\arctan\left(x\right)\big) \tan\left(\arctan\left(\frac{1}{x} \right)\right)}=\frac{x+(1/x)}{0}, $$ which is undefined. Since the tangent function is undefined, in $[-\pi, \pi]$, if and only if its argument is $\pm \pi/2$, then $f(x)=\pm \pi/2$. It's easy to see that if $x<0$, then $\arctan(x)<0$, hence $f(x)=-\pi/2$ and viceversa.

I have found a few other solution to this problem, but I wanted to know if this one is logically acceptable.

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    $\begingroup$ Thumbs up from my side! Will be curious to see if someone can find a problem with this. $\endgroup$ – Deepak Gupta Sep 11 '15 at 6:04
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    $\begingroup$ Cute.You can re-phrase it to say "if tan f(x) exists then it equals (x+1/x)/0 ,therefore it doesn't exist,therefore |f(x)|= pi/2" $\endgroup$ – DanielWainfleet Sep 11 '15 at 6:28
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    $\begingroup$ Reminder to Deepak Gupta .It's \infty , not \infinity in LaTeX. $\endgroup$ – DanielWainfleet Sep 11 '15 at 6:30
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    $\begingroup$ Looks good to me. The sequence of equations starting with $\tan(f(x))$ assumes (at first) that $\tan(f(x))$ is defined, then proves by contradiction that it is not. You could say "assume $\tan(f(x))$ is defined" before those equations, but I don't think you really need to. The assumption is implied by the fact that you wrote an equation. $\endgroup$ – David K Sep 13 '15 at 16:28
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    $\begingroup$ I think there's a problem with the use of the formula related to $\tan(x+y)$ because this formula is valid when everything is defined. I think it's better to infere $f(x)=\pm\frac\pi2$ only by showing that $1-\tan\arctan x\tan\arctan\frac1x=0$. $\endgroup$ – Mohsen Shahriari Sep 13 '15 at 22:36
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Whatever you have done is correct and nothing is wrong there.You could get the same as follows also: Note that : $\tan^{-1}x=\tan^{-1}1/x -\pi$, when $x\lt0$...(A) and is equal to $\tan^{-1}1/x$ when $x\gt0$..(B)

PROOF:I am proving for $x$ being negative.For +ve $x$ it's quite easy to prove.

Suppose, $y=-z$, ($z$ is positive ), hen $\cot^{-1}(y)=\cot^{-1}(-z)=\pi -\cot^{-1}(z)$ Now you might be knowing :$\cot^{-1}(z)=\tan^{-1}(1/z)$ for positive $z$, hence $\cot^{-1}(-z)=\pi - \tan^{-1}(1/z)$. Now substitue $z=-y$, then use $\tan^{-1}(-1/y)=-\tan^{-1}(1/y)$ and you get the result.

Now you may argue: why $\cot^{-1}(-z)=\pi-\cot^{-1}z$?

Well, it's because :Let $z=\cot\theta$, $0<\theta<\pi$[principal branch of $\cot$]. So $-z=\cot(\pi-\theta)$ , Note that $\pi- \theta$ is also in between $0$ and $\pi$. So you can define $\cot^{-1}(-z)$, which in this case will be: $\pi -\theta$, Now put $\theta=\cot^{-1}(z)$, Hence proved.

So coming to: $\tan^{-1}x +\tan^{-1}1/x=I$, say

So $I=\tan^{-1}x +\cot^{-1}x=\pi/2$, when $x\gt0$ [by (B)]

And $I=\tan^{-1}x +\cot^{-1}x-\pi=\pi/2-\pi=-\pi/2$, when $x \lt0$ [by (A)]

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    $\begingroup$ Do you see a typographical difference between $\tan^{-1}x$=$\tan^{-1}1/x$ -$\pi$ and $\tan^{-1}x=\tan^{-1}1/x -\pi$? The latter is proper usage. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 19 '15 at 0:48
  • $\begingroup$ @MichaelHardy,both are the same.What do you actually want to say?Please explain. $\endgroup$ – Koro Sep 19 '15 at 1:34
  • $\begingroup$ @MichaelHardy,oh! now I see.Thanks for information! $\endgroup$ – Koro Sep 19 '15 at 1:38
  • $\begingroup$ @MichaelHardy,anyways what is wrong with the first usage??Please explain. $\endgroup$ – Koro Sep 19 '15 at 1:39
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    $\begingroup$ Look at it. In the first example, the minus sign before $\pi$ looks like a hypen rather than a minus sign and lacks spacing to its left and right, thus $1/x\text{-}\pi$ instead of $1/x-\pi$. The "equals" sign is also to small for the font size and lacks spacing no the left and right, thus: $\tan^{-1}x$=$\tan^{-1}$ etc. instead of $\tan^{-1}x=\tan$ etc. ${}\qquad{}$ $\endgroup$ – Michael Hardy Sep 19 '15 at 2:05
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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, $\alpha+\beta={\pi\over2}$ for any $x$.

enter image description here

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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 $.

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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

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

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The tangent function should be defined to take the value $\infty$ at $\pm\pi/2$, 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 $\pm\pi/2$.

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{\huge\text{?}}=\ \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} \tag 1 \\[10pt] & {} = \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}. \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 $\cos\alpha\cos\beta$, getting this: $$ \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta} = \frac{\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}}{1 - \frac{\sin\alpha}{\cos\alpha}\cdot\frac{\sin\beta}{\cos\beta}} \tag 3 $$ 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 $\cos\alpha\ne0\ne\cos\beta$. 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 $\cos\alpha=0$. Then

$$ \frac\pi2 = - \frac\pi2 = \alpha+\beta = \frac\pi2+\beta = -\frac\pi2 + \beta $$ and we're done.

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You can't compute $\tan\theta$ when $\theta=\pm\frac\pi2$, but you can compute $\cos\theta$ instead. For $a,b\in \mathbb{R}$ with $ab\ne 1$ 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 $$ \cos^2(f(x))=\lim_{a\to x,b\to x^{-1}}\cos^2(\arctan(a)+\arctan(b))=\lim_{a\to x,b\to x^{-1}}\frac{(1-ab)^2}{(1-ab)^2+(ab)^2}=0, $$ i.e. $f(x)=\pm\frac\pi2$. Since $f$ is continuous on $(-\infty,0)$ and on $(0,\infty)$, we deduce that $$ f(x)=f(-1)=2\arctan(-1)=-\frac\pi2 \quad \forall x<0, $$ and $$ f(x)=-f(-x)=\frac\pi2 \quad \forall x>0 $$


Added The function $f: x\mapsto f(x)=\arctan(x)+\arctan(x^{-1})$ defined and differentiable of $\mathbb{R}\setminus\{0\}$. For every $x\ne 0$ 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 $$ f(1)=2\arctan(1)=2\dfrac{\pi}{4}=\dfrac{\pi}{2},\quad f(-1)=-f(1)=-\dfrac{\pi}{2}, $$ it follows that $$ \arctan(x)+\arctan\left(\dfrac{1}{x}\right)=\begin{cases}-\dfrac{\pi}{2} &\text{ if } x<0\\ \dfrac{\pi}{2} &\text{ if } x>0 \end{cases}. $$

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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. $

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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 $x<0$.

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Here's a fun route - suppose

\begin{eqnarray}\arctan(p)+\arctan(q)+\arctan(r)+\arctan(s)&=\pi \\ \implies \arg[(1+ip)(1+iq)(1+ir)(1+is)]&=\pi \\ \implies p+q+r+s-pqs-pqt-qst-pst&=0\end{eqnarray}

which is solved by e.g. $(p,q,r,s)=(x,\frac{1}{x},y,\frac{1}{y})$. We can set \begin{equation}2[\arctan(x)+\arctan(\frac{1}{x})]=\pi\end{equation}

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