92. \( \cos \theta=-\frac{1}{4}, \quad \tan \theta>0 \)

Given the conditions: \[ \cos \theta = -\frac{1}{4}, \quad \tan \theta > 0 \] Let's analyze these to determine the quadrant in which angle \(\theta\) lies and find relevant trigonometric values. ### 1. Determining the Quadrant - **\(\cos \theta = -\frac{1}{4}\)**: - The cosine of an angle is negative in **Quadrants II** and **III**. - **\(\tan \theta > 0\)**: - The tangent of an angle is positive when both sine and cosine have the **same sign**. - Since \(\cos \theta\) is negative, for \(\tan \theta\) to be positive, \(\sin \theta\) must also be negative. **Conclusion**: Both sine and cosine are negative, which occurs in **Quadrant III**. ### 2. Finding \(\sin \theta\) Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] \[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \left(-\frac{1}{4}\right)^2 = 1 - \frac{1}{16} = \frac{15}{16} \] \[ \sin \theta = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4} \] (The negative sign is chosen because \(\sin \theta\) is negative in Quadrant III.) ### 3. Finding \(\tan \theta\) \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{\sqrt{15}}{4}}{-\frac{1}{4}} = \sqrt{15} \] ### Summary - **Quadrant**: III - **\(\sin \theta\)**: \(-\frac{\sqrt{15}}{4}\) - **\(\tan \theta\)**: \(\sqrt{15}\) If you have a specific aspect of the problem you'd like to explore further, feel free to ask!