4 If \( \cos \theta=-\frac{1}{5} \) and \( \theta \) lies in Quadrant II, what is the value of \( \tan \theta \) ? \( \begin{array}{llll}\text { 1) } \frac{3}{4} & \text { 2) } \frac{4}{3} & \text { 3) }-\frac{3}{4} & \text { 4) }-\frac{4}{3} \end{array} \)

Given that \( \cos \theta = -\frac{1}{5} \) and \( \theta \) is in Quadrant II, we can determine \( \tan \theta \) using the following steps: 1. **Understanding Trigonometric Ratios in Quadrant II:** - In Quadrant II, cosine (\( \cos \)) is negative, and sine (\( \sin \)) is positive. - Tangent (\( \tan \)) is the ratio of sine to cosine. Thus, since \( \sin \) is positive and \( \cos \) is negative in Quadrant II, \( \tan \theta \) will be negative. 2. **Using the Pythagorean Identity:** - We know that \( \sin^2 \theta + \cos^2 \theta = 1 \). - Given \( \cos \theta = -\frac{1}{5} \), we find \( \sin \theta \): \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(-\frac{1}{5}\right)^2 = 1 - \frac{1}{25} = \frac{24}{25} \] \[ \sin \theta = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5} \] 3. **Calculating the Tangent:** \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} = -2\sqrt{6} \] However, the options provided are rational numbers, suggesting a possible intended simplification or a different approach based on standard triangle ratios. Typically, in a 3-4-5 triangle scaled appropriately: - If \( \cos \theta = -\frac{3}{5} \) (a common value in such triangles), then: \[ \tan \theta = \frac{4}{-3} = -\frac{4}{3} \] Since \(-\frac{4}{3}\) is among the provided options and aligns with standard triangle ratios, it's likely the intended answer. **Answer:** 4) \(-\frac{4}{3}\)