Find the exact value of the indicated trigonometric function of \( \theta \). \[ \begin{array}{l}\text { 11) } \sin \theta=\frac{1}{2}, \sec \theta<0 \\ \text { and } \tan \theta\end{array} \]

To find the value of \(\tan \theta\) given that \(\sin \theta = \frac{1}{2}\) and \(\sec \theta < 0\), we start by identifying the possible quadrant for \(\theta\). Since \(\sin \theta\) is positive and \(\sec \theta\) (which is the reciprocal of \(\cos \theta\)) is negative, it indicates that \(\theta\) is in the second quadrant. In the second quadrant, we know that \(\sin \theta = \frac{1}{2}\) corresponds to an angle of \(\frac{\pi}{6}\) (or \(30^\circ\)), but in the second quadrant, the reference angle would be \(\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}\). To find \(\tan \theta\), we use the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\). Since \(\sin \theta = \frac{1}{2}\), we can find \(\cos \theta\) using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \left(\frac{1}{2}\right)^2 + \cos^2 \theta = 1 \] \[ \frac{1}{4} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{1}{4} = \frac{3}{4} \] So, \(\cos \theta = -\sqrt{\frac{3}{4}} = -\frac{\sqrt{3}}{2}\) (since \(\sec \theta < 0\) implies \(\cos \theta < 0\)). Now, we can compute \(\tan \theta\): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \] Thus, the exact value of \(\tan \theta\) is \[ \boxed{-\frac{\sqrt{3}}{3}}. \]