Find the five remaining trigonometic functions of $\theta$.
$$\sec \theta =\frac{8}{7},\sin \theta <0$$
Complete the following tablens
$$\begin{array}{l}\sin \theta =\\ \cos \theta =\\ \tan \theta =\end{array}$$
$$\begin{array}{l}\csc \theta =◻\\ \sec \theta =\frac{8}{7}\\ \cot \theta =◻\end{array}$$
(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: $\boxed{{\displaystyle \sin \theta =-\frac{\sqrt{15}}{8},\cos \theta =\frac{7}{8},\tan \theta =-\sqrt{15},\csc \theta =-\frac{8}{\sqrt{15}},\sec \theta =\frac{8}{7},\cot \theta =-\frac{1}{\sqrt{15}}}}$