Answer
$$\cos\theta=-\frac{\sqrt 5}{3}$$
$$\tan\theta=-\frac{2\sqrt 5}{5}$$
$$\cot\theta=-\frac{\sqrt 5}{2}$$
$$\csc\theta=\frac{3}{2}$$
$$\sec\theta=\frac{-3\sqrt 5}{5}$$
Work Step by Step
$$\sin\theta=\frac{2}{3}\hspace{1.5cm}\theta\hspace{0.1cm}\text{in quadrant II}$$
1) Reciprocal Identities:
$$\csc\theta=\frac{1}{\sin\theta}$$
$$\csc\theta=\frac{1}{\frac{2}{3}}=\frac{3}{2}$$
2) Pythagorean Identities:
$$\cot^2\theta+1=\csc^2\theta$$
$$\cot^2\theta=\csc^2\theta-1$$
$$\cot^2\theta=(\frac{3}{2})^2-1=\frac{9}{4}-1=\frac{5}{4}$$
$$\cot\theta=\pm\frac{\sqrt5}{2}$$
3) Quotient Identities:
$$\cot\theta=\frac{\cos\theta}{\sin\theta}$$
We know that $\theta$ is in quadrant II, where $\sin\theta\gt0$ but $\cos\theta\lt0$. Therefore, with the above identity, we can conclude that $\cot\theta\lt0$.
$$\cot\theta=-\frac{\sqrt 5}{2}$$
$$\frac{\cos\theta}{\sin\theta}=-\frac{\sqrt 5}{2}$$
$$\cos\theta=(-\frac{\sqrt 5}{2})\times\sin\theta=(-\frac{\sqrt 5}{2})\times\frac{2}{3}=-\frac{\sqrt 5}{3}$$
4) Reciprocal Identities:
$$\cot\theta=\frac{1}{\tan\theta}$$
$$\tan\theta=\frac{1}{\cot\theta}=\frac{1}{-\frac{\sqrt 5}{2}}=-\frac{2}{\sqrt 5}=-\frac{2\sqrt 5}{5}$$
Also, $$\sec\theta=\frac{1}{\cos\theta}=\frac{1}{-\frac{\sqrt 5}{3}}=-\frac{3}{\sqrt 5}=-\frac{3\sqrt 5}{5}$$
