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Calculator says the answer is $180^\circ$. But from the formula $$\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1} \left(\frac{x+y}{1-xy}\right)$$ we can also say that the answer is $0$. Which is correct?

edit: If I use "radian" mode it says 0, in degree mode it is saying 180, why?

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  • $\begingroup$ If $\tan^{-1}(x) = \arctan(x)$ then the calculator is wrong. $\arctan(-x) = -\arctan(x)$ for all $x$. $\endgroup$ Oct 26 '15 at 6:27
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    $\begingroup$ What does your calculator say for just $\tan^{-1}(-3)$ all by itself? And for $\tan^{-1}(3)$? If you are not getting $\approx\pm71.565^{\circ}$ for these, then your calculator is not using the standard meaning of $\tan^{-1}$. $\endgroup$ Oct 26 '15 at 6:32
  • $\begingroup$ The standard function arctan takes values in $(-\pi/2,\pi/2)$.Your calculator seems to be unaware of this. $\endgroup$ Oct 26 '15 at 7:17
  • $\begingroup$ It's possible that when your calculator switches to degree mode, it's been programmed to assume you want positive angles to be reported whenever possible, so it takes $90\lt \tan^{-1}u\lt180$ for $u\lt0$. What make and model of calculator are you using? $\endgroup$ Oct 28 '15 at 16:45
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Since $\arctan x$ is an odd function, $\arctan(-3) = -\arctan 3$, so the sum is zero.

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  • $\begingroup$ But the calculator says, answer is 180 $\endgroup$
    – Shafaet
    Oct 26 '15 at 7:42
  • $\begingroup$ Your calculator is wrong. The range for $\arctan x$ is $(-\frac{\pi}{2},\frac{\pi}{2})$. The only way you can return a value of $\pi$ (or $180^{\circ}$) is to calculate $\lim_{x \to \infty} \arctan x - \arctan (-x) = \lim_{x \to \infty} \arctan x + \arctan x = \lim_{x \to \infty} 2\arctan x$. You can simulate something close to this by keying in huge arguments like $\arctan(1E6) - \arctan(-1E6)$ while in degrees mode and seeing what you get (it'll be very close to $180^{\circ}$). On the other hand, the calculation you mentioned in the question should definitely work out to be zero. $\endgroup$
    – Deepak
    Oct 26 '15 at 8:11
  • $\begingroup$ @Shafaet I can't figure out how that can be so. $\endgroup$
    – Deepak
    Oct 26 '15 at 9:38
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To propose an alternative to existing answers: since

$$\tan(x)=\tan(x+\pi)$$

holds for all $x\in\mathbb R$, one can well say that $\tan^{-1}(y)$ is only defined up to an integer multiple of $\pi$. In complex analysis that's the usual view, as far as I understand it.

If you want to obtain a single value as the result, however, you have to specify one branch of that multi-valued function. And the regular convention is something like $\tan^{-1}(y)\in(-\frac\pi2,\frac\pi2]$ (with the value $\frac\pi2$ only reachable if you allow $\infty$ as a valid input). In that setup, your calculator is just wrong, since apparently it's picking the wrong branch at some point.

One could imagine that passing large values to the arctangent function you could end up in a situation where it's numerically unclear which side of the branch cut a given value lies on. But $3$ should be sufficuently far from infinity that this shouldn't be confusing your calculator.

Perhaps the calculator is trying to simplify the term symbolically before evaluating it numerically. And perhaps its handling of branch cuts isn't good enough in this setup, causing it to symbolically make some simplification which would correspond to a branch other than the main branch for the given numeric input.

It's hard to know more details without even knowing the model of the calulator, whether it's sophisticated enough to attempt symbolic simplification and dumb enough to choose wrong branches in doing so.

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  1. Why don't you tell us what your calculator says for $\tan^{-1}(-3)$?

  2. Surely $180$ is the result in "degree" mode, not "radian" mode? If you insist otherwise, then I insist on photographic evidence.

  3. To answer your thrice-repeated question: your calculator is wrong.

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  • $\begingroup$ Sorry, I said the opposite. $\endgroup$
    – Shafaet
    Oct 28 '15 at 16:33

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