find the exact values of the five trigonometric ratios not given.\n22) \\( \\csc \\theta = \\frac { \\sqrt {…

Explanation:

Step1: Find (\sin\theta)

Since (\csc\theta=\frac{\sqrt{29}}{5}), and (\csc\theta = \frac{1}{\sin\theta}), then (\sin\theta=\frac{5}{\sqrt{29}}=\frac{5\sqrt{29}}{29})

Step2: Determine the quadrant

Given (\cos\theta<0) and (\sin\theta>0) (because (\sin\theta=\frac{5\sqrt{29}}{29}>0)), (\theta) is in the second quadrant.

Step3: Use the Pythagorean identity (\sin^{2}\theta+\cos^{2}\theta = 1)

Substitute (\sin\theta=\frac{5\sqrt{29}}{29}) into the identity: (\left(\frac{5\sqrt{29}}{29}\right)^{2}+\cos^{2}\theta=1) (\frac{25\times29}{841}+\cos^{2}\theta=1) (\frac{25}{29}+\cos^{2}\theta=1) (\cos^{2}\theta=1 - \frac{25}{29}=\frac{29 - 25}{29}=\frac{4}{29}) Since (\cos\theta<0) in the second quadrant, (\cos\theta=-\frac{2\sqrt{29}}{29})

Step4: Calculate (\tan\theta)

(\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{5\sqrt{29}}{29}}{-\frac{2\sqrt{29}}{29}}=-\frac{5}{2})

Step5: Calculate (\sec\theta)

(\sec\theta=\frac{1}{\cos\theta}=-\frac{\sqrt{29}}{2})

Step6: Calculate (\cot\theta)

(\cot\theta=\frac{\cos\theta}{\sin\theta}=\frac{-\frac{2\sqrt{29}}{29}}{\frac{5\sqrt{29}}{29}}=-\frac{2}{5})

Answer:

(\sin\theta=\frac{5\sqrt{29}}{29}), (\cos\theta =-\frac{2\sqrt{29}}{29}), (\tan\theta=-\frac{5}{2}), (\sec\theta=-\frac{\sqrt{29}}{2}), (\cot\theta=-\frac{2}{5})