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?

There's a bit of a discrepancy between the page title and your actual question.
Conventionally, $\tan^{-1}(0)$ is defined as 0.
However, the individual terms $\tan^{-1} 1$, $\tan^{-1} 2$, and $\tan^{-1} 3$ are all positive numbers, so their sum can't be zero.

*

*$\tan^{-1} 1 = \frac{\pi}{4}$

*$\tan^{-1} 2$ and $\tan^{-1} 3$ are hard to express exactly, but each must be between $\tan \sqrt{3} = \frac{\pi}{3}$ and $\tan^{-1} \infty = \frac{\pi}{2}$
So, you can determine that $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3$ must have a value somewhere in the interval $[\frac{\pi}{4} + \frac{\pi}{3} + \frac{\pi}{3}, \frac{\pi}{4} + \frac{\pi}{2} + \frac{\pi}{2}] = [\frac{11}{12} \pi, \frac{5}{4}\pi]$.  And it happens that $\pi$ is the only answer choice that falls within that range.
