92. \( \cos \theta=-\frac{1}{4}, \quad \tan \theta>0 \)
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To find the angle \( \theta \) given that \( \cos \theta = -\frac{1}{4} \) and \( \tan \theta > 0 \), we first note that since \( \cos \theta \) is negative, \( \theta \) must be in the second or third quadrant. However, \( \tan \theta > 0 \) indicates that \( \theta \) is in the third quadrant, where both sine and cosine are negative. Using the cosine value, we can find the sine using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \): \[ \sin^2 \theta + \left(-\frac{1}{4}\right)^2 = 1 \implies \sin^2 \theta + \frac{1}{16} = 1 \] \[ \sin^2 \theta = 1 - \frac{1}{16} = \frac{15}{16} \implies \sin \theta = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4} \] The sign is negative because \( \theta \) is in the third quadrant. Now, we have \( \cos \theta = -\frac{1}{4} \) and \( \sin \theta = -\frac{\sqrt{15}}{4} \). If we need to find \( \theta \), we can use the tangent relation: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{\sqrt{15}}{4}}{-\frac{1}{4}} = \sqrt{15}. \] Thus, you can find the reference angle \( \theta = \pi + \tan^{-1}(\sqrt{15}) \) for the exact value. The final answer for \( \theta \) in radians is: \[ \theta = \pi + \tan^{-1}(\sqrt{15}). \]
