Step 1 :We are given that \(\sin \theta=\frac{\sqrt{5}}{7}\) and \(\theta\) is in quadrant II. We can use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find \(\cos \theta\). Since \(\theta\) is in quadrant II, \(\cos \theta\) will be negative.
Step 2 :Using the Pythagorean identity, we find that \(\cos \theta = -\sqrt{1 - \sin^2 \theta} = -\sqrt{1 - \left(\frac{\sqrt{5}}{7}\right)^2} = -\frac{\sqrt{44}}{7}\).
Step 3 :The reciprocal of \(\sin \theta\) is \(\csc \theta\), so we can find \(\csc \theta\) by taking the reciprocal of \(\sin \theta\). This gives us \(\csc \theta = \frac{1}{\sin \theta} = \frac{7}{\sqrt{5}}\).
Step 4 :Similarly, the reciprocal of \(\cos \theta\) is \(\sec \theta\), so we can find \(\sec \theta\) by taking the reciprocal of \(\cos \theta\). This gives us \(\sec \theta = \frac{1}{\cos \theta} = -\frac{7}{\sqrt{44}}\).
Step 5 :Finally, \(\tan \theta\) is the ratio of \(\sin \theta\) to \(\cos \theta\), so we can find \(\tan \theta\) by dividing \(\sin \theta\) by \(\cos \theta\). This gives us \(\tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{\sqrt{5}}{\sqrt{44}}\).
Step 6 :\(\boxed{\cos \theta = -\frac{\sqrt{44}}{7}}\)
Step 7 :\(\boxed{\csc \theta = \frac{7}{\sqrt{5}}}\)
Step 8 :\(\boxed{\sec \theta = -\frac{7}{\sqrt{44}}}\)
Step 9 :\(\boxed{\tan \theta = -\frac{\sqrt{5}}{\sqrt{44}}}\)