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

Given that \( \cot(\theta) = \frac{8}{15} \) and \( \cos(\theta) < 0 \), we can determine \( \sec(\theta) \) as follows: 1. **Understanding the Ratios:** - \( \cot(\theta) = \frac{\text{adjacent}}{\text{opposite}} = \frac{8}{15} \). - This suggests that in a right triangle, the adjacent side is 8 units and the opposite side is 15 units. 2. **Determining the Hypotenuse:** - Using the Pythagorean theorem: \[ \text{hypotenuse} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17. \] 3. **Considering the Quadrant:** - \( \cos(\theta) < 0 \) implies that \( \theta \) is in either the second or third quadrant. - Since \( \cot(\theta) = \frac{8}{15} > 0 \), and cotangent is positive only in the first and third quadrants, \( \theta \) must be in the **third quadrant**. - In the third quadrant, both adjacent and opposite sides are negative. 4. **Calculating \( \cos(\theta) \):** - \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{-8}{17} \). 5. **Finding \( \sec(\theta) \):** - \( \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{8}{17}} = -\frac{17}{8} \). **Answer:** \( \sec (\theta) = -\dfrac{17}{8} \)