find the remaining five trigonometric functions of ( \theta ).\n\n( sin \theta=\frac{2}{3}, \theta ) in…

Explanation:

Step1: Find (\csc\theta)

Since (\csc\theta=\frac{1}{\sin\theta}), and (\sin\theta = \frac{2}{3}), then (\csc\theta=\frac{1}{\frac{2}{3}}=\frac{3}{2}).

Step2: Find (\cos\theta)

Using the identity (\sin^{2}\theta+\cos^{2}\theta = 1), so (\cos\theta=-\sqrt{1-\sin^{2}\theta}) (because (\theta) is in quadrant II, (\cos\theta<0)). Substitute (\sin\theta=\frac{2}{3}), we get (\cos\theta=-\sqrt{1 - (\frac{2}{3})^{2}}=-\sqrt{1-\frac{4}{9}}=-\sqrt{\frac{5}{9}}=-\frac{\sqrt{5}}{3}).

Step3: Find (\sec\theta)

Since (\sec\theta=\frac{1}{\cos\theta}), and (\cos\theta=-\frac{\sqrt{5}}{3}), then (\sec\theta=\frac{1}{-\frac{\sqrt{5}}{3}}=-\frac{3}{\sqrt{5}}=-\frac{3\sqrt{5}}{5}).

Step4: Find (\tan\theta)

Since (\tan\theta=\frac{\sin\theta}{\cos\theta}), substitute (\sin\theta=\frac{2}{3}) and (\cos\theta=-\frac{\sqrt{5}}{3}), we get (\tan\theta=\frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}}=-\frac{2}{\sqrt{5}}=-\frac{2\sqrt{5}}{5}).

Step5: Find (\cot\theta)

Since (\cot\theta=\frac{1}{\tan\theta}), and (\tan\theta=-\frac{2\sqrt{5}}{5}), then (\cot\theta=\frac{1}{-\frac{2\sqrt{5}}{5}}=-\frac{5}{2\sqrt{5}}=-\frac{\sqrt{5}}{2}).

Answer:

(\cos\theta =-\frac{\sqrt{5}}{3}), (\tan\theta =-\frac{2\sqrt{5}}{5}), (\csc\theta=\frac{3}{2}), (\sec\theta =-\frac{3\sqrt{5}}{5}), (\cot\theta =-\frac{\sqrt{5}}{2})