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.

  • 2
    $\begingroup$ wrong, even without the squares $\endgroup$ – Jean-Claude Arbaut May 7 '14 at 12:07
  • $\begingroup$ @Jean-ClaudeArbaut are you sure it's wrong? $\endgroup$ – Venus May 7 '14 at 12:11
  • 1
    $\begingroup$ Take any calculator, and try with any value of $x\in[0,\pi]$ except $\pi/4$ and $3\pi/4$. $\endgroup$ – 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

$$\tan^{-1} y = \frac{\pi}{2} - \cot^{-1} z.$$


$$\tan(\tan^{-1} y) = \tan\left(\frac{\pi}{2} - \cot^{-1} z\right).$$

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.