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The equation is: $\arctan 3 + 2\arctan2 = \pi + arccot 3$

They go on and assign $\arctan 3 = \theta$ and $\arctan2 = \phi$. Therefore, $\frac{\pi}{4}< \theta < \frac{\pi}{2}$, same for $\phi$. Which I follow.

Then they use $\tan(A+B)$ formula to show that that $\tan(\theta+2\phi)=\frac{1}{3}$. Which implies that $\arctan \frac{1}{3} = \theta + 2\phi$, but apparently not so. As it implies $\pi + \arctan \frac{1}{3} = \theta + 2\phi$.

And there I am lost. I can see that when you combine the inequalities you get $$\pi+\frac{\pi}{4} < \theta + 2\phi < \pi + \frac{\pi}{2}$$

And then they go on to show that $$\theta + 2\phi = \pi + \arctan \frac{1}{3} = \pi + arccot 3$$

I do not understand where that $\pi$ is coming from there...

Naz
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    When $\tan x = \tan y$, then it does not follow that $x = y$, but only that $x-y = k\pi$ for some $k\in \mathbb{Z}$. Here the inequalities show that $k = 1$ is the right integer. – Daniel Fischer Aug 07 '15 at 16:30
  • If you were to write up this a bit more formally I would go for your answer. As I indeed know that the general solution to $\tan \theta = s$ is $PV + n\Pi$. Therefore $\tanx = \tan(x+n\Pi)$. But will the inequality always show a multiple of $\pi$? I have a tendency to ask before thinking. – Naz Aug 07 '15 at 16:44

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First of all, $\arctan x+\text{arccot}x=\dfrac\pi2$ (Proof)

So, the problem reduces to $\arctan 2+\arctan3=\dfrac{3\pi}4$

Now from this or Ex$\#5$ of Page $\#276$ of this, $$\tan^{-1}x+\tan^{-1}y=\begin{cases} \tan^{-1}\dfrac{x+y}{1-xy} &\mbox{if } xy<1 \\\pi+ \tan^{-1}\dfrac{x+y}{1-xy} & \mbox{if } xy> 1. \end{cases} $$

Finally, the principal value of $\displaystyle\tan$ lies $\displaystyle\in\left[-\frac\pi2,\frac\pi2\right]$

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