19. arctan\\frac{\\sqrt{3}}{3}

Explanation:

Step1: Recall the definition of arctangent function

The function (y = \arctan(x)) is the inverse function of (y=\tan(x)) with the domain (x\in(-\frac{\pi}{2},\frac{\pi}{2})). We need to find an angle (\theta\in(-\frac{\pi}{2},\frac{\pi}{2})) such that (\tan\theta=\frac{\sqrt{3}}{3}).

Step2: Use the values of tangent function for special angles

We know that (\tan\frac{\pi}{6}=\frac{\sin\frac{\pi}{6}}{\cos\frac{\pi}{6}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}), and (\frac{\pi}{6}\in(-\frac{\pi}{2},\frac{\pi}{2}))

Answer:

(\frac{\pi}{6})