find the remaining five trigonometic functions of \\( \\theta \\).\n\\( \\sin \\theta = \\frac { 2 } { 5 }…

Explanation:

Step1: Find (\csc\theta)

Since (\csc\theta=\frac{1}{\sin\theta}), and (\sin\theta = \frac{2}{5}), then (\csc\theta=\frac{5}{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}{5}), we get (\cos\theta=-\sqrt{1 - (\frac{2}{5})^{2}}=-\sqrt{1-\frac{4}{25}}=-\sqrt{\frac{21}{25}}=-\frac{\sqrt{21}}{5}).

Step3: Find (\sec\theta)

Since (\sec\theta=\frac{1}{\cos\theta}), then (\sec\theta=-\frac{5}{\sqrt{21}}=-\frac{5\sqrt{21}}{21}).

Step4: Find (\tan\theta)

Since (\tan\theta=\frac{\sin\theta}{\cos\theta}), substitute (\sin\theta=\frac{2}{5}) and (\cos\theta =-\frac{\sqrt{21}}{5}), we get (\tan\theta=\frac{\frac{2}{5}}{-\frac{\sqrt{21}}{5}}=-\frac{2}{\sqrt{21}}=-\frac{2\sqrt{21}}{21}).

Step5: Find (\cot\theta)

Since (\cot\theta=\frac{1}{\tan\theta}), then (\cot\theta=-\frac{\sqrt{21}}{2}).

Answer:

(\cos\theta=-\frac{\sqrt{21}}{5}), (\tan\theta =-\frac{2\sqrt{21}}{21}), (\csc\theta=\frac{5}{2}), (\sec\theta=-\frac{5\sqrt{21}}{21}), (\cot\theta=-\frac{\sqrt{21}}{2})