Is $\tan^{-1}(0)=\pi$ or is $\tan^{-1}(0)=0$?

Question

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?

Answer 1

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.

Answer 2

It is not true that $$x:=\tan^{-1}a+\tan^{-1}b+\tan^{-1}c =\tan^{-1}\frac{a+b+c-abc}{1-ab-ac-bc}$$

It is true that $$\tan x=\frac{a+b+c-abc}{1-ab-ac-bc}.$$

But $\tan^{-1}(\tan x)\neq x$ in general.

Answer 3

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 $(c)$.

Answer 4

People usually define $\tan^{-1}$ so that its domain is $(-\infty, \infty)$ and its range is $(-\pi / 2, \pi / 2)$. This means that $\tan^{-1}(0) = 0$, not $\pi$.