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..

Solutions Collecting From Web of "Verify the identity: $\tan^{-1} x +\tan^{-1} (1/x) = \pi /2$"

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
$$
\alpha , \beta, \alpha + \beta \neq \frac\pi 2 \mod \pi
$$

An easy, mostly graphical proof: $\tan\alpha=x$, $\tan\beta=\frac1x$, and $\alpha+\beta=\frac\pi2$.

Triangle

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.

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 $0<\arctan(1/x)<\pi/2$.

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 $\frac{\pi}{2}$

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}

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 $z=\pi/2-y$.