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.
Best Answer
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 $x$.