Find the exact value of each of the remaining trigonometric functions of $\theta$. Rationalize denominators when applicable. $\sec \theta=-10$, given that $\sin \theta>0$

Step 1 ：Given that \(\sec \theta = -10\), we know that \(\cos \theta = -1/10\) because secant is the reciprocal of cosine.

Step 2 ：Since \(\sin \theta > 0\) and \(\cos \theta < 0\), we know that \(\theta\) is in the second quadrant.

Step 3 ：We can use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find \(\sin \theta\). Solving for \(\sin \theta\), we get \(\sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - (-1/10)^2} = \sqrt{99}/10\).

Step 4 ：Using the definitions of the other trigonometric functions in terms of sine and cosine, we find that \(\tan \theta = \sin \theta / \cos \theta = -\sqrt{99}\), \(\csc \theta = 1/\sin \theta = 10/\sqrt{99}\), and \(\cot \theta = 1/\tan \theta = -1/\sqrt{99}\).

Step 5 ：\(\boxed{\text{Final Answer: } \sin \theta = \sqrt{99}/10, \cos \theta = -1/10, \tan \theta = -\sqrt{99}, \csc \theta = 10/\sqrt{99}, \cot \theta = -1/\sqrt{99}}\)