Suppose that \( \beta \) is an angle with \( \cos \beta=\frac{6}{11} \) and \( \beta \) is not in the fourth quadrant. Compute the exact value of \( \tan \beta \). You do not have to rationalize the denominator. \( \tan \beta= \)

To find \( \tan \beta \) given that \( \cos \beta = \frac{6}{11} \) and that \( \beta \) is not in the fourth quadrant, we can use the Pythagorean identity: \[ \sin^2 \beta + \cos^2 \beta = 1. \] First, we calculate \( \sin^2 \beta \): \[ \sin^2 \beta = 1 - \cos^2 \beta. \] Substituting \( \cos \beta = \frac{6}{11} \): \[ \cos^2 \beta = \left( \frac{6}{11} \right)^2 = \frac{36}{121}. \] Now, substituting this into the Pythagorean identity: \[ \sin^2 \beta = 1 - \frac{36}{121} = \frac{121}{121} - \frac{36}{121} = \frac{85}{121}. \] Next, we take the square root to find \( \sin \beta \): \[ \sin \beta = \sqrt{\sin^2 \beta} = \sqrt{\frac{85}{121}} = \frac{\sqrt{85}}{11}. \] Since \( \beta \) is not in the fourth quadrant, \( \sin \beta \) is positive. Thus, we have: \[ \sin \beta = \frac{\sqrt{85}}{11}. \] Now we can compute \( \tan \beta \): \[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{\sqrt{85}}{11}}{\frac{6}{11}} = \frac{\sqrt{85}}{6}. \] Thus, the exact value of \( \tan \beta \) is: \[ \boxed{\frac{\sqrt{85}}{6}}. \]