Question 13, 1.4.77 HW Score: 80\%, 12 of Part 4 of 5 Pointss 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. \( \cos \theta=-\frac{\sqrt{39}}{26} \) (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 sinswer box to complete your choice. A. \( \tan \theta=-\frac{7 \sqrt{3}}{3} \) (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. \( \csc \theta= \) \( \square \) (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 -

To find the exact value of \( \csc \theta \) given that \( \cot \theta = -\frac{\sqrt{3}}{7} \) and \( \theta \) is in Quadrant II, we'll follow these steps: ### Step 1: Understand the Given Information - **Quadrant II**: In this quadrant, sine is positive, and cosine is negative. - **\( \cot \theta = \frac{\cos \theta}{\sin \theta} = -\frac{\sqrt{3}}{7} \)**: This implies that \( \cos \theta \) is negative and \( \sin \theta \) is positive. ### Step 2: Set Up the Right Triangle Let’s consider a right triangle where: - **Adjacent side** (to angle \( \theta \)): \( -\sqrt{3} \) (negative because cosine is negative in Quadrant II) - **Opposite side**: \( 7 \) ### Step 3: Find the Hypotenuse Using the Pythagorean theorem: \[ \text{Hypotenuse} = \sqrt{(\text{Adjacent})^2 + (\text{Opposite})^2} = \sqrt{(-\sqrt{3})^2 + 7^2} = \sqrt{3 + 49} = \sqrt{52} = 2\sqrt{13} \] ### Step 4: Calculate \( \sin \theta \) \[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{7}{2\sqrt{13}} \] Rationalizing the denominator: \[ \sin \theta = \frac{7}{2\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}} = \frac{7\sqrt{13}}{26} \] ### Step 5: Find \( \csc \theta \) \[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{7\sqrt{13}}{26}} = \frac{26}{7\sqrt{13}} \] Rationalizing the denominator: \[ \csc \theta = \frac{26}{7\sqrt{13}} \times \frac{\sqrt{13}}{\sqrt{13}} = \frac{26\sqrt{13}}{91} = \frac{2\sqrt{13}}{7} \] ### Final Answer **A.** \( \csc \theta = \frac{2\sqrt{13}}{7} \)