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  1. MHB Apprentice

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    #1
    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.

  2. Perseverance
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    #2
    For $x>0$, $\arctan(x)$ and $\arctan\left(\frac1x\right)$ are complementary angles.

  3. MHB Seeker
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    #3
    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[right] {$x$}
    (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 $\arctan x + \arctan \frac 1x = \frac\pi 2$.

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