Part 5 of 5 Points: 0 of 1 Find the exact value of each of the remaining trigonometric functions of \( \theta \). Rationalize denominators when applicable. \( \cot \theta=-\frac{\sqrt{3}}{7} \), given that \( \theta \) is in quadrant II B. The function is undefined. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \boldsymbol{\operatorname { t a n }} \theta=-\frac{7 \sqrt{3}}{3} \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \csc \theta=\frac{2 \sqrt{13}}{7} \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. \( \sec \theta= \) \( \square \) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) B. The function is undefined. View an example Get more help -
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To find the remaining trigonometric functions based on \( \cot \theta = -\frac{\sqrt{3}}{7} \) in quadrant II, we start with the definition of cotangent. Since \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can set \( \cos \theta = -\sqrt{3} \) and \( \sin \theta = 7 \) (but we need to adjust these values into a valid triangle using the Pythagorean identity). Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we find: 1. Calculate the hypotenuse: \[ r = \sqrt{(-\sqrt{3})^2 + 7^2} = \sqrt{3 + 49} = \sqrt{52} = 2\sqrt{13}. \] 2. Then, we can determine: \[ \sin \theta = \frac{7}{2\sqrt{13}}, \quad \cos \theta = -\frac{\sqrt{3}}{2\sqrt{13}}. \] 3. Finally, we calculate: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{7}{\sqrt{3}} \rightarrow \tan \theta = -\frac{7 \sqrt{3}}{3} \] (after rationalizing the denominator), \[ \csc \theta = \frac{1}{\sin \theta} = \frac{2\sqrt{13}}{7}, \quad \sec \theta = \frac{1}{\cos \theta} = -\frac{2\sqrt{13}}{\sqrt{3}} = -\frac{2\sqrt{39}}{3} \] (after rationalizing the denominator). So, the answers are: 1. For \( \tan \theta \), the choice is A: \( \tan \theta = -\frac{7 \sqrt{3}}{3} \). 2. For \( \csc \theta \), the choice is A: \( \csc \theta = \frac{2\sqrt{13}}{7} \). 3. For \( \sec \theta \), the choice is A: \( \sec \theta = -\frac{2\sqrt{39}}{3} \).