Answer
$$\sin\theta=-\frac{3\sqrt{10}}{10}$$
Work Step by Step
$$\cot\theta=-\frac{1}{3}$$
According to the Pythagorean Identities: $$1+\cot^2\theta=\csc^2\theta=\frac{1}{\sin^2\theta}$$
That means $$\sin^2\theta=\frac{1}{1+\cot^2\theta}=\frac{1}{1+\Big(-\frac{1}{3}\Big)^2}=\frac{1}{1+\frac{1}{9}}=\frac{1}{\frac{10}{9}}=\frac{9}{10}$$
So, $$\sin\theta=\pm\frac{3}{\sqrt{10}}$$
Another given information shows that $\theta$ is in quadrant IV, which means $\sin\theta\lt0$.
Therefore, $$\sin\theta=-\frac{3}{\sqrt{10}}=-\frac{3\sqrt{10}}{10}$$
