Solve tan⁻¹x + tan⁻¹3x = π/4 by taking the tangent of both sides…

Step 1. The quadratic. $\dfrac{x + 3x}{1 - 3x^2} = 1$ gives $4x = 1 - 3x^2$, so $3x^2 + 4x - 1 = 0$ and $x = \dfrac{-2 \pm \sqrt{7}}{3}$. The roots are $0.2153$ and $-1.5486$.

Step 2. The first root. For $x = 0.2153$ the product $x \cdot 3x = 0.139 < 1$, so the sum rule applies, and the sum is $\tan^{-1}1 = \dfrac{\pi}{4}$. It is a real solution.

Step 3. The other root. For $x = -1.5486$ the product is $7.19 > 1$ and $x < 0$. The sum is $-\pi + \dfrac{\pi}{4} = -\dfrac{3\pi}{4}$, which is $-0.75\pi$. The tangent of $-\dfrac{3\pi}{4}$ is also $1$, but the angle is not $\dfrac{\pi}{4}$. So this root is false.