33. \( \sec \theta=-3, \quad \tan \theta>0 \)

To solve for \( \theta \) given that \( \sec \theta = -3 \) and \( \tan \theta > 0 \), we can start by recalling that since \( \sec \theta = \frac{1}{\cos \theta} \), it follows that \( \cos \theta = -\frac{1}{3} \). The given condition of \( \tan \theta > 0 \) indicates that \( \theta \) must be in the third quadrant (where sine and cosine values have opposite signs). Next, we can find \( \sin \theta \) using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Substituting \( \cos \theta = -\frac{1}{3} \): \[ \sin^2 \theta + \left(-\frac{1}{3}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{1}{9} = 1 \] \[ \sin^2 \theta = 1 - \frac{1}{9} = \frac{8}{9} \] This means: \[ \sin \theta = -\sqrt{\frac{8}{9}} = -\frac{2\sqrt{2}}{3} \] (We take the negative value because sine is also negative in the third quadrant.) Now we can find \( \theta \) using the cosine value. The reference angle can be calculated as: \[ \theta = \pi + \cos^{-1}\left(-\frac{1}{3}\right) \] Thus, we also conclude that the angle can be expressed in radians or degrees, maintaining the conditions specified. Finally, to summarize: - \( \sec \theta = -3 \) leads to \( \cos \theta = -\frac{1}{3} \) and \( \sin \theta = -\frac{2\sqrt{2}}{3} \) in the third quadrant.