Answer
$$\sin\theta=-\frac{2}{5}$$
$$\cos\theta=-\frac{\sqrt{21}}{5}$$
$$\tan\theta=\frac{2\sqrt{21}}{21}$$
$$\cot\theta=\frac{\sqrt{21}}{2}$$
$$\sec\theta=-\frac{5\sqrt{21}}{21}$$
Work Step by Step
$$\csc\theta=-\frac{5}{2}\hspace{1.5cm}\theta\hspace{0.1cm}\text{in quadrant III}$$
1) Reciprocal Identities:
$$\csc\theta=\frac{1}{\sin\theta}$$
$$\sin\theta=\frac{1}{\csc\theta}=\frac{1}{-\frac{5}{2}}=-\frac{2}{5}$$
2) Pythagorean Identities:
$$\csc^2\theta=\cot^2\theta+1$$
$$\cot^2\theta=\csc^2\theta-1=(-\frac{5}{2})^2-1=\frac{25}{4}-1=\frac{21}{4}$$
$$\cot\theta=\pm\frac{\sqrt{21}}{2}$$
Also,
$$\cos^2\theta=1-\sin^2\theta=1-(-\frac{2}{5})^2=1-\frac{4}{25}=\frac{21}{25}$$
$$\cos\theta=\pm\frac{\sqrt{21}}{5}$$
We know that $\theta$ is in quadrant IV, where both $\sin\theta\lt0$ and $\cos\theta\lt0$. Therefore, $$\cos\theta=-\frac{\sqrt{21}}{5}$$
3) Quotient Identities:
$$\cot\theta=\frac{\cos\theta}{\sin\theta}$$
We already have $\sin\theta\lt0$ and $\cos\theta\lt0$.That means, with the above identity, $\cot\theta\gt0$
$$\cot\theta=\frac{\sqrt{21}}{2}$$
Also, we have
$$\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{-\frac{2}{5}}{-\frac{\sqrt{21}}{5}}=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}$$
4) Reciprocal Identities:
$$\sec\theta=\frac{1}{\cos\theta}=\frac{1}{-\frac{\sqrt{21}}{5}}=-\frac{5}{\sqrt{21}}=-\frac{5\sqrt{21}}{21}$$
