Find the five remaining trigonometic functions of $\theta$. \[ \sec \theta=\frac{8}{7}, \sin \theta<0 \] Complete the following tablens \[ \begin{array}{l} \sin \theta= \\ \cos \theta= \\ \tan \theta= \end{array} \] \[ \begin{array}{l} \csc \theta=\square \\ \sec \theta=\frac{8}{7} \\ \cot \theta=\square \end{array} \] (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.)

Step 1 ：Given that \(\sec \theta=\frac{8}{7}\), and \(\sin \theta<0\)

Step 2 ：Since \(\sec \theta\) is the reciprocal of \(\cos \theta\), we can find \(\cos \theta\) by taking the reciprocal of \(\sec \theta\), so \(\cos \theta = \frac{7}{8}\)

Step 3 ：Since we are in the fourth quadrant where cosine is positive and sine is negative, we can find \(\sin \theta\) using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\), so \(\sin \theta = -\sqrt{1 - \cos^2 \theta} = -\sqrt{1 - \left(\frac{7}{8}\right)^2} = -\frac{\sqrt{15}}{8}\)

Step 4 ：\(\tan \theta\) is the sine divided by the cosine, so \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\sqrt{15}/8}{7/8} = -\sqrt{15}\)

Step 5 ：\(\csc \theta\) is the reciprocal of the sine, so \(\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\sqrt{15}/8} = -\frac{8}{\sqrt{15}}\)

Step 6 ：\(\cot \theta\) is the reciprocal of the tangent, so \(\cot \theta = \frac{1}{\tan \theta} = \frac{1}{-\sqrt{15}} = -\frac{1}{\sqrt{15}}\)

Step 7 ：Final Answer: \(\boxed{\sin \theta= -\frac{\sqrt{15}}{8}, \cos \theta= \frac{7}{8}, \tan \theta= -\sqrt{15}, \csc \theta= -\frac{8}{\sqrt{15}}, \sec \theta= \frac{8}{7}, \cot \theta= -\frac{1}{\sqrt{15}}}\)