Find \( \sec (\theta) \) if \( \cot (\theta)=\frac{8}{15} \) and \( \cos (\theta)<0 \). Answer exactly. \( \sec (\theta)=\square \)

To find \( \sec(\theta) \), we start with the given information that \( \cot(\theta) = \frac{8}{15} \). Since \( \cot(\theta) = \frac{1}{\tan(\theta)} \), this means \( \tan(\theta) = \frac{15}{8} \). Now, we consider a right triangle where the opposite side to angle \( \theta \) is \( 15 \) and the adjacent side is \( 8 \). The hypotenuse \( r \) can be calculated using the Pythagorean theorem: \[ r = \sqrt{(8^2 + 15^2)} = \sqrt{64 + 225} = \sqrt{289} = 17. \] Next, we find \( \cos(\theta) \) and \( \sec(\theta) \): \[ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{8}{17}. \] However, since we need \( \cos(\theta) < 0 \), \( \theta \) must be in the second quadrant. Thus, \[ \cos(\theta) = -\frac{8}{17}. \] Now, we can find \( \sec(\theta) \): \[ \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{8}{17}} = -\frac{17}{8}. \] Thus, the answer is: \[ \sec(\theta) = -\frac{17}{8}. \]