Is $\tan^{-1}(0)=\pi$ or is $\tan^{-1}(0)=0$? This question came in the Dhaka University admission exam 2006-7
Q) The value of $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$ is -
(a) $0$
(b) $\frac{\pi}{2}$
(c) $\pi$
(d) $2\pi$
My attempt:
$$\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$$
$$=\tan^{-1}\frac{1+2+3-1\cdot2\cdot3}{1-1\cdot2-2\cdot3-1\cdot3}$$
$$=\tan^{-1}\frac{0}{-10}$$
$$=\tan^{-1}0$$
Now, as no range is mentioned, I should pick the angle that is within the principal range of $\tan^{-1}$: (a). (c) and (d) are also acceptable solutions, but they don't fall within the principal range. So, I'll go with (a).
However, my question bank says that $\tan^{-1}0=\pi$, so the answer is (b). I don't understand their reasoning.
Which is the correct option?

Let's find the value of $\arctan 2 + \arctan 3$ first...
Using the formula $$\arctan x + \arctan y = \arctan \left(\dfrac {x+y}{1-xy}\right)$$ we have $$\arctan 2 + \arctan 3 = \arctan \left(\dfrac {2+3}{1-2 \cdot 3}\right) \\ = \arctan \dfrac {5}{-5} \\ = \arctan (-1) \\ = \dfrac {3 \pi}{4}$$ but since $\arctan 1 = \dfrac {\pi}{4}$, $$\arctan 1 + \arctan 2 + \arctan 3 = \dfrac {\pi}{4} + \dfrac {3 \pi}{4} = \pi$$ which is answer $(c)$.
