Find the exact value of each of the remaining trigonometric functions of \( \theta \). Rationalize denominators when applicable. \( \sec \theta=-5 \), given that \( \sin \theta>0 \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \sin \theta=\square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined.

Since \( \sec \theta = -5 \), we know \( \cos \theta = -\frac{1}{5} \). Now, using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can find \( \sin \theta \). Substituting \( \cos \theta \): \[ \sin^2 \theta + \left(-\frac{1}{5}\right)^2 = 1 \] \[ \sin^2 \theta + \frac{1}{25} = 1 \] \[ \sin^2 \theta = 1 - \frac{1}{25} = \frac{25}{25} - \frac{1}{25} = \frac{24}{25} \] Now, taking the square root: \[ \sin \theta = \sqrt{\frac{24}{25}} = \frac{\sqrt{24}}{5} = \frac{2\sqrt{6}}{5} \] Since we know \( \sin \theta > 0 \), we have: A. \( \sin \theta = \frac{2\sqrt{6}}{5} \) Now, let's find the exact values of the remaining trigonometric functions. 1. **\( \tan \theta = \frac{\sin \theta}{\cos \theta} \)** \[ \tan \theta = \frac{\frac{2\sqrt{6}}{5}}{-\frac{1}{5}} = -2\sqrt{6} \] 2. **\( \csc \theta = \frac{1}{\sin \theta} = \frac{5}{2\sqrt{6}} = \frac{5\sqrt{6}}{12} \) (after rationalizing)** 3. **\( \cot \theta = \frac{1}{\tan \theta} = -\frac{1}{2\sqrt{6}} = -\frac{\sqrt{6}}{12} \) (after rationalizing)** 4. **\( \cos \theta = -\frac{1}{5} \) already stated.** Finally, the values for all the trigonometric functions are: - \( \sin \theta = \frac{2\sqrt{6}}{5} \) - \( \cos \theta = -\frac{1}{5} \) - \( \tan \theta = -2\sqrt{6} \) - \( \csc \theta = \frac{5\sqrt{6}}{12} \) - \( \sec \theta = -5 \) - \( \cot \theta = -\frac{\sqrt{6}}{12} \) So, to complete the answer: A. \( \sin \theta = \frac{2\sqrt{6}}{5} \)