1. **State the problem:** Given $\cos \theta = -\frac{2}{5}$ and $\sin \theta > 0$, find the quadrant of $\theta$ and the values of $\sin \theta$, $\tan \theta$, $\sec \theta$, $\csc \theta$, and $\cot \theta$. 2. **Determine the quadrant:** Since $\cos \theta$ is negative and $\sin \theta$ is positive, $\theta$ lies in Quadrant II where cosine is negative and sine is positive. 3. **Find $\sin \theta$:** Use the Pythagorean identity: $$\sin^2 \theta + \cos^2 \theta = 1$$ Substitute $\cos \theta = -\frac{2}{5}$: $$\sin^2 \theta + \left(-\frac{2}{5}\right)^2 = 1$$ $$\sin^2 \theta + \frac{4}{25} = 1$$ $$\sin^2 \theta = 1 - \frac{4}{25} = \frac{25}{25} - \frac{4}{25} = \frac{21}{25}$$ Since $\sin \theta > 0$, $$\sin \theta = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5}$$ 4. **Find $\tan \theta$:** $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{21}}{5}}{-\frac{2}{5}} = -\frac{\sqrt{21}}{2}$$ 5. **Find $\sec \theta$:** $$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{2}{5}} = -\frac{5}{2}$$ 6. **Find $\csc \theta$:** $$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{\sqrt{21}}{5}} = \frac{5}{\sqrt{21}} = \frac{5\sqrt{21}}{21}$$ 7. **Find $\cot \theta$:** $$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{-\frac{\sqrt{21}}{2}} = -\frac{2}{\sqrt{21}} = -\frac{2\sqrt{21}}{21}$$ **Final answers:** - Quadrant: II - $\sin \theta = \frac{\sqrt{21}}{5}$ - $\tan \theta = -\frac{\sqrt{21}}{2}$ - $\sec \theta = -\frac{5}{2}$ - $\csc \theta = \frac{5\sqrt{21}}{21}$ - $\cot \theta = -\frac{2\sqrt{21}}{21}$