Verify the identity: $\tan^{-1} x +\tan^{-1} (1/x) = \pi /2$ Verify the identity: $\tan^{-1} x + \tan^{-1} (1/x) = \frac\pi 2, x > 0$
$$\alpha= \tan^{-1} x$$
$$\beta = \tan^{-1} (1/x)$$
$$\tan \alpha = x$$
$$\tan \beta = 1/x$$
$$\tan^{-1}[\tan(\alpha + \beta)]$$
$$\tan^{-1}\left
[{\tan\alpha + \tan\beta\over 1 - \tan\alpha \tan\beta}
\right]$$
$$\tan^{-1}\left[
{x + 1/x\over 1- x/x }\right]$$
$$\tan^{-1}\left[{x + (1/x)\over 0} \right]$$
I can't find out what I'm doing wrong..
 A: An easy, mostly graphical proof: $\tan\alpha=x$, $\tan\beta=\frac1x$, and $\alpha+\beta=\frac\pi2$.

The reason you get a division by zero in the argument of arctan is that $\displaystyle\lim_{\varphi\to\frac\pi2}\tan\varphi=\pm\infty\approx\tfrac10$. So, in very informal notation, you could say that $\tan^{-1}(\infty)=\tfrac\pi2$, and that your calculation in a way make sense.
A: You're basically trying to compute $\tan(\pi/2)$, which doesn't exist.
If you set $\beta=\arctan(1/x)$, then $\tan\beta=1/x$, that is
$$
x=\cot\beta=\tan\left(\frac{\pi}{2}-\beta\right)
$$
Therefore
$$
\arctan x=\arctan\tan\left(\frac{\pi}{2}-\beta\right)=\frac{\pi}{2}-\beta
$$
by the hypothesis that $x>0$, so that $0<\arctan(1/x)<\pi/2$.
A: Assume $x>0$, then
\begin{align}&\tan^{-1} x +\tan^{-1} \dfrac1x
\\\\=&\tan^{-1} x +\tan^{-1}\dfrac1{\tan\tan^{-1} x}
\\\\=&\tan^{-1} x +\tan^{-1}\cot \tan^{-1} x
\\\\=&\tan^{-1} x +\tan^{-1}\tan(\dfrac{\pi}2- \tan^{-1}x)
\\\\=&\tan^{-1} x +\dfrac{\pi}2- \tan^{-1}x\qquad\qquad\qquad
\left(\because\text{for $x>0$,}\;\dfrac{\pi}2- \tan^{-1}x\in\left(0,\dfrac{\pi}2\right)\right)
\\\\=&\dfrac{\pi}2.
\end{align}
A: One may also use complex numbers: We are multiplying two complex numbers with argument $\frac{1}{x}$ and $x$.
So, we desire to show that $\arg((1 + ix)(x + i)) = \frac{\pi}{2}$
We expand the product to get $(x^2 + 1)i$ -- since there is no real part and the imaginary part is $> 0$, the argument is $\frac{\pi}{2}$
A: Yet another possibility:
You want $y+z$ for $y,z$ satisfying $\sin y / \cos y=\cos z/\sin z$.  Since $\sin y=\cos(\pi/2-y)$ it follows that $z=\pi/2-y$.
A: Hint: 
When you want to prove that something smooth is constant, use derivatives.
details:
if $f(x) = \arctan x + \arctan\frac 1x$ then
$$
f'(x) = \frac 1{1+x^2} + \frac 1{1+\left(\frac 1x\right)^2}\times \left(-\frac{1}{x^2}\right) =0
$$
then $f(x) = f(1) = 2\arctan 1 = \frac\pi 2$ on the interval $\{x>0\}$.
The problem of your method is that the formula you are using is true only when
$$
\alpha , \beta, \alpha + \beta \neq \frac\pi 2 \mod \pi
$$
A: In addition to that which was exposed, I want to say the following:
$\cot(\tan^{-1}x ) = \tan(\pi/2 - \tan^{-1}x)$
Let's prove it:
First, observe that $\cot(\tan^{-1}x) = \cot(\tan^{-1}x + \cot^{-1}x -\cot^{-1}x ) = \cot(\pi/2 -\cot^{-1}x) $ (I)
since $\tan^{-1}x+\cot^{-1}x = \pi/2 $ (II)
But, observe that from equation (I) and also that $\cot(\pi/2 - x) = \tan x$, that $\cot(\pi/2 -\cot^{-1}x) = \tan(\cot^{-1}x).$
Now, we conclude (II).
