# Trigonometry Question: How to prove $\arctan(\cot^2(x)) + \mathrm{arccot}(\tan^2 (x)) = \frac{\pi}{2}$?

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.

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wrong, even without the squares –  Jean-Claude Arbaut May 7 '14 at 12:07
@Jean-ClaudeArbaut are you sure it's wrong? –  Venus May 7 '14 at 12:11
Take any calculator, and try with any value of $x\in[0,\pi]$ except $\pi/4$ and $3\pi/4$. –  Jean-Claude Arbaut May 7 '14 at 12:13

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 $\cot^2 x = \tan^2 x$.

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Thank you David. –  Venus May 8 '14 at 14:13