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..

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}
