Question 11, 1.4,69 Points: 0 of 1 Find the exact value of \( \tan \theta \), given that \( \sin \theta=-\frac{1}{6} \) and \( \theta \) is in quadrant IV. Rationalize denominators when applicable. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \boldsymbol{\operatorname { t a n }} \theta= \) \( \square \) (Simplify your answer. Type an exact answer, using radicals as needed. Type an integer or a fraction.) B. The function is undefined.

To find the exact value of \( \tan \theta \), we start with the given \( \sin \theta = -\frac{1}{6} \). Since \( \theta \) is in quadrant IV, we know that \( \cos \theta \) is positive. We can use the Pythagorean identity, \( \sin^2 \theta + \cos^2 \theta = 1 \). Calculating \( \sin^2 \theta \): \[ \sin^2 \theta = \left(-\frac{1}{6}\right)^2 = \frac{1}{36} \] Now, using the identity: \[ \frac{1}{36} + \cos^2 \theta = 1 \] \[ \cos^2 \theta = 1 - \frac{1}{36} = \frac{36}{36} - \frac{1}{36} = \frac{35}{36} \] Thus, \( \cos \theta = \sqrt{\frac{35}{36}} = \frac{\sqrt{35}}{6} \) (we take the positive root since \( \theta \) is in the IV quadrant). Now we can determine \( \tan \theta \) using the definition: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{1}{6}}{\frac{\sqrt{35}}{6}} = -\frac{1}{\sqrt{35}} \] To rationalize the denominator: \[ \tan \theta = -\frac{1}{\sqrt{35}} \cdot \frac{\sqrt{35}}{\sqrt{35}} = -\frac{\sqrt{35}}{35} \] So, the final answer is: A. \( \boldsymbol{\operatorname { t a n }} \theta= -\frac{\sqrt{35}}{35} \)