How to prove the identity:
$$\arctan(\cot^2(x)) + \mathrm{arccot}(\tan^2 (x)) = \frac{\pi}{2}?$$
I've been doing and googling for hours to prove that identity but I can't and I only found this video, still I can't figure it out. Any help would be appreciated. Thanks in advance.
Let's examine this by using $y$ in place of the quantity $\cot^2 x$ and $z$ in place of $\tan^2 x$ in your formula. Then what we're asked to show was that
$$\tan^{-1} y + \cot^{-1} z = \frac{\pi}{2}.$$
Let's ask a slightly different question: when can this equation be true? The equation above implies that
$$\tan^{-1} y = \frac{\pi}{2} - \cot^{-1} z.$$
Therefore
$$\tan(\tan^{-1} y) = \tan\left(\frac{\pi}{2} - \cot^{-1} z\right).$$
But $\tan\left(\frac{\pi}{2} - \theta\right) = \cot \theta$, so the equation above is simply saying that
$$\tan(\tan^{-1} y) = \cot(\cot^{-1} z),$$
or in other words,
$$y = z.$$
Therefore, recalling how we defined $y$ and $z$, the "identity" we were supposed to prove is true only when $\cot^2 x = \tan^2 x$.