Prove that arctan(1) + arctan(2) + arctan(3) = pi

Question

Prove that $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi$ .

Solution — Step by Step

The key identity:

\tan^{-1}(a) + \tan^{-1}(b) = \begin{cases} \tan^{-1}\left(\dfrac{a+b}{1-ab}\right) & \text{if } ab < 1 \\ \pi + \tan^{-1}\left(\dfrac{a+b}{1-ab}\right) & \text{if } ab > 1, \, a > 0 \\ -\pi + \tan^{-1}\left(\dfrac{a+b}{1-ab}\right) & \text{if } ab > 1, \, a < 0 \end{cases}

The condition $ab > 1$ requires the extra $\pm\pi$ term.

Let’s first add $\tan^{-1}(2) + \tan^{-1}(3)$ .

Here $a = 2$ , $b = 3$ , so $ab = 6 > 1$ and $a > 0$ .

\tan^{-1}(2) + \tan^{-1}(3) = \pi + \tan^{-1}\left(\frac{2+3}{1-6}\right) = \pi + \tan^{-1}\left(\frac{5}{-5}\right) = \pi + \tan^{-1}(-1)

= \pi + \left(-\frac{\pi}{4}\right) = \frac{3\pi}{4}

Now we have:

\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \frac{\pi}{4} + \frac{3\pi}{4} = \frac{4\pi}{4} = \pi

\boxed{\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi}

Why This Works

The reason the formula adds $\pi$ when $ab > 1$ comes from the range restriction of $\tan^{-1}$ : it outputs values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ . When the actual sum exceeds this range (which happens when $ab > 1$ and both $a, b > 0$ ), we must add $\pi$ to correct for the “wrap-around.”

Geometrically: the three angles $\frac{\pi}{4}$ , $\tan^{-1}(2) \approx 63.4°$ , and $\tan^{-1}(3) \approx 71.6°$ add up to $180°$ — this is a pure coincidence of arithmetic for these specific values of 1, 2, 3.

Alternative Method — Tangent of the Sum

We can verify: $\tan(\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)) = \tan(\pi) = 0$ .

First: $\tan(\tan^{-1}(1) + \tan^{-1}(2)) = \frac{1+2}{1-1\times 2} = \frac{3}{-1} = -3$

So $\tan^{-1}(1) + \tan^{-1}(2) = \pi + \tan^{-1}(-3) = \pi - \tan^{-1}(3)$

Therefore: $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) = \pi - \tan^{-1}(3) + \tan^{-1}(3) = \pi$ ✓

Common Mistake

Applying the addition formula without checking the condition $ab > 1$ . When $ab < 1$ , the sum of two arctans stays within the principal range and no $\pi$ adjustment is needed. When $ab > 1$ (with both positive), we must add $\pi$ . For $a = 2$ , $b = 3$ : $ab = 6 > 1$ , so we add $\pi$ to the formula result. Skipping this correction gives $\frac{\pi}{4} + (-\frac{\pi}{4}) = 0$ , completely wrong.

Question

Solution — Step by Step

Why This Works

Alternative Method — Tangent of the Sum

Common Mistake

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