MHB Equation involving the inverse tangent function

[karseme]

I need to prove that:

$ \arctan{\dfrac{1}{x}}=\dfrac{\pi}{2}- \arctan{x}, \forall x>0$.

Now, I assumed $\arctan{\dfrac{1}{x}}=\arccot{x}$. So, I've tried to do this:

$\cot{y}=x \implies y=arccot{x} \\ \tan{y}=\dfrac{1}{\cot{y}}=\dfrac{1}{x} \implies y=\arctan{\dfrac{1}{x}} \\ \implies \arccot{x}=\arctan{\dfrac{1}{x}}$. I've tried to put in some numbers and it seems that it workes for every real number.

Also, $\tan{(\dfrac{\pi}{2}-y)}=\cot{y}=x \implies \dfrac{\pi}{2}-y=\arctan{x} \land y=\arccot{x} \\ \implies \arctan{x}+\arccot{x}=\dfrac{\pi}{2}$, which also works for every real number. But, why is it then when you plug in $\arccot{x}=\arctan{\dfrac{1}{x}}$ in the second equation, it doesn't work for every x. But, the first equation and the second equation work for every real number but their combination doesn't. I know that my approach wasn't that good anyway, but I didn't know what else to do to prove this.

 

[Greg]

For $x>0$, $\arctan(x)$ and $\arctan\left(\frac1x\right)$ are complementary angles.

 

[I like Serena]

Hi karseme! ;)

Consider the following right triangle:
\begin{tikzpicture}[font=\large]
\draw[ultra thick, blue]
(0,0) node[above right,xshift=10] {$\alpha$} -- node[below] {$1$}
(4,0) -- node

{$x$}
(4,3) node[below left,yshift=-6] {$\beta$} -- cycle;
\draw[blue] (4,0) rectangle +(-0.3,0.3);
\end{tikzpicture}

From the definition of $\tan$ we have $\tan\alpha=\frac x 1$ and $\tan\beta=\frac 1 x$.
From the angle sum of a triangle we know that $\alpha + \beta=\frac\pi 2$.
Therefore $\arctan x + \arctan \frac 1x = \frac\pi 2$. ​