1. **State the problem:** Given $\cos \theta = \frac{6}{10} = 0.6$ and $\tan \theta < 0$, find the values of $\sin \theta$, $\tan \theta$, $\csc \theta$, $\sec \theta$, and $\cot \theta$. 2. **Recall the Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ This allows us to find $\sin \theta$ when $\cos \theta$ is known. 3. **Find $\sin \theta$:** $$\sin^2 \theta = 1 - \cos^2 \theta = 1 - (0.6)^2 = 1 - 0.36 = 0.64$$ $$\sin \theta = \pm \sqrt{0.64} = \pm 0.8$$ 4. **Determine the sign of $\sin \theta$ using $\tan \theta < 0$:** Since $\tan \theta = \frac{\sin \theta}{\cos \theta}$ and $\cos \theta = 0.6 > 0$, for $\tan \theta$ to be negative, $\sin \theta$ must be negative. Therefore, $\sin \theta = -0.8$. 5. **Calculate $\tan \theta$:** $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-0.8}{0.6} = -\frac{4}{3} \approx -1.3333$$ 6. **Calculate $\csc \theta$ (reciprocal of $\sin \theta$):** $$\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-0.8} = -\frac{5}{4} = -1.25$$ 7. **Calculate $\sec \theta$ (reciprocal of $\cos \theta$):** $$\sec \theta = \frac{1}{\cos \theta} = \frac{1}{0.6} = \frac{5}{3} \approx 1.6667$$ 8. **Calculate $\cot \theta$ (reciprocal of $\tan \theta$):** $$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{-\frac{4}{3}} = -\frac{3}{4} = -0.75$$ **Final answers:** - $\sin \theta = -0.8$ - $\tan \theta = -\frac{4}{3}$ - $\csc \theta = -\frac{5}{4}$ - $\sec \theta = \frac{5}{3}$ - $\cot \theta = -\frac{3}{4}$