(c) If \( \sin \theta=\frac{12}{13} \) and \( \pi / 2<\theta<3 \pi / 2 \), find \( \begin{array}{lll}\text { i. } \tan \theta & \text { ii. } \cos ^{2} \theta-\sin ^{2} \theta & \text { iii. } 13 \cos \theta+5 \tan \theta\end{array} \)

**Step 1.** Use the Pythagorean identity to find \(\cos \theta\). Given \[ \sin^2 \theta + \cos^2 \theta = 1 \quad \text{and} \quad \sin \theta = \frac{12}{13}, \] we have \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169}. \] Taking the square root gives \[ \cos \theta = \pm \frac{5}{13}. \] Since \(\frac{\pi}{2} < \theta < \frac{3\pi}{2}\) (second or third quadrant) and \(\sin \theta\) is positive, \(\theta\) must be in the second quadrant where cosine is negative. Therefore, \[ \cos \theta = -\frac{5}{13}. \] --- **Step 2.** Find \(\tan \theta\). Using the definition \[ \tan \theta = \frac{\sin \theta}{\cos \theta}, \] we substitute the known values: \[ \tan \theta = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5}. \] --- **Step 3.** Compute \(\cos^2 \theta - \sin^2 \theta\). We have \[ \cos^2 \theta - \sin^2 \theta = \left(-\frac{5}{13}\right)^2 - \left(\frac{12}{13}\right)^2 = \frac{25}{169} - \frac{144}{169} = -\frac{119}{169}. \] --- **Step 4.** Evaluate \(13 \cos \theta + 5 \tan \theta\). Substitute the values of \(\cos \theta\) and \(\tan \theta\): \[ 13 \cos \theta + 5 \tan \theta = 13\left(-\frac{5}{13}\right) + 5\left(-\frac{12}{5}\right) = -5 - 12 = -17. \] --- **Final Answers:** i. \(\tan \theta = -\frac{12}{5}\) ii. \(\cos^2 \theta - \sin^2 \theta = -\frac{119}{169}\) iii. \(13 \cos \theta + 5 \tan \theta = -17\)