1. The problem is to prove the identity $$\arctan(1) + \arctan(2) + \arctan(3) = \pi$$. 2. We use the formula for the sum of arctangents: $$\arctan(a) + \arctan(b) = \arctan\left(\frac{a+b}{1 - ab}\right)$$ if $ab < 1$, otherwise we add $\pi$. 3. First, calculate $$\arctan(1) + \arctan(2)$$: $$\arctan(1) + \arctan(2) = \arctan\left(\frac{1+2}{1 - 1\times 2}\right) = \arctan\left(\frac{3}{1 - 2}\right) = \arctan\left(\frac{3}{-1}\right) = \arctan(-3)$$. Since $1 \times 2 = 2 > 1$, we add $\pi$: $$\arctan(1) + \arctan(2) = \arctan(-3) + \pi$$. 4. Now add $$\arctan(3)$$: $$\arctan(-3) + \arctan(3) + \pi$$. Using the sum formula again: $$\arctan(-3) + \arctan(3) = \arctan\left(\frac{-3 + 3}{1 - (-3)(3)}\right) = \arctan\left(\frac{0}{1 + 9}\right) = \arctan(0) = 0$$. 5. Therefore: $$\arctan(1) + \arctan(2) + \arctan(3) = 0 + \pi = \pi$$. 6. This completes the proof that $$\arctan(1) + \arctan(2) + \arctan(3) = \pi$$.