Problem

The key insight is that the three angles of a triangle must sum to $\pi$, so you need to find what $\tan^{-1} 2 + \tan^{-1} 3$ equals, then subtract from $\pi$. Use the tangent addition formula for inverse tangents: when $xy < 1$, $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$. Here, since $2 \cdot 3 = 6 > 1$, you're in the case where the sum exceeds $\frac{\pi}{2}$, which requires careful handling of the formula's domain.

The key insight is that the three angles of a triangle must sum to $\pi$, so you need to find what $\tan^{-1} 2 + \tan^{-1} 3$ equals, then subtract from $\pi$. Use the tangent addition formula for inverse tangents: when $xy < 1$, $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$. Here, since $2 \cdot 3 = 6 > 1$, you're in the case where the sum exceeds $\frac{\pi}{2}$, which requires careful handling of the formula's domain.