1. **State the problem:** Given $\csc \theta = -1.45$ and $\theta$ is in quadrant III, find $\cot \theta$. 2. **Recall identities:** - $\csc \theta = \frac{1}{\sin \theta}$ - $\cot \theta = \frac{\cos \theta}{\sin \theta}$ - In quadrant III, both $\sin \theta$ and $\cos \theta$ are negative. 3. **Find $\sin \theta$:** $$\sin \theta = \frac{1}{\csc \theta} = \frac{1}{-1.45} = -\frac{1}{1.45} \approx -0.6897$$ 4. **Use Pythagorean identity to find $\cos \theta$:** $$\sin^2 \theta + \cos^2 \theta = 1$$ $$\cos^2 \theta = 1 - \sin^2 \theta = 1 - (-0.6897)^2 = 1 - 0.4757 = 0.5243$$ $$\cos \theta = -\sqrt{0.5243} \approx -0.7243$$ (negative because quadrant III) 5. **Calculate $\cot \theta$:** $$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{-0.7243}{-0.6897} = \frac{\cancel{-}0.7243}{\cancel{-}0.6897} \approx 1.05$$ **Final answer:** $$\cot \theta \approx 1.05$$