10-seconds challenge-5

Let $\begin{cases} \alpha = \tan^{-1} 1 & \Rightarrow \tan \alpha = 1 \\ \beta = \tan^{-1} 2 & \Rightarrow \tan \beta = 2 \\ \gamma = \tan^{-1} 3 & \Rightarrow \tan \gamma = 3 \end{cases} \quad \Rightarrow \alpha + \beta + \gamma = n \pi$

Now, we have:

$\begin{aligned} \tan (\alpha + \beta) & = \frac{1+2}{1-(1)(2)} = - 3 \\ \tan (\alpha + \beta + \gamma) & = \frac{-3+3}{1-(-3)(3)} = 0 \\ \Rightarrow \tan n \pi & = 0 \quad \quad \small \color{#3D99F6}{\text{For } \alpha, \beta, \gamma < \frac{\pi}{2}} \\ \Rightarrow n & = \boxed{1} \end{aligned}$

$\tan^{-1}(x)+\tan^{-1}(y)=\pi+\tan^{-1}(\dfrac{x+y}{1-xy})$ if $xy>1$ and $\tan^{-1}(1)=\dfrac{\pi}{4}$

Therefore, $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\dfrac{\pi}{4} + \pi+ \tan^{-1}(\dfrac{2+3}{1-6})=\dfrac{\pi}{4}+\pi-\dfrac{\pi}{4}=\boxed\pi$