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

To solve for \( \tan \theta \), \( \cos^2 \theta - \sin^2 \theta \), and \( 13 \cos \theta + 5 \tan \theta \) given that \( \sin \theta = \frac{12}{13} \) and \( \pi/2 < \theta < 3\pi/2 \), we start by determining \( \cos \theta \). Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have: \[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \] Thus, \[ \cos \theta = -\sqrt{\frac{25}{169}} = -\frac{5}{13} \] (here we take the negative root because \(\theta\) is in the third quadrant, where cosine is negative). Now, we can find \( \tan \theta \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{-\frac{5}{13}} = -\frac{12}{5} \] Next, we calculate \( \cos^2 \theta - \sin^2 \theta \): \[ \cos^2 \theta - \sin^2 \theta = \frac{25}{169} - \frac{144}{169} = -\frac{119}{169} \] Finally, we compute \( 13 \cos \theta + 5 \tan \theta \): \[ 13 \cos \theta + 5 \tan \theta = 13\left(-\frac{5}{13}\right) + 5\left(-\frac{12}{5}\right) = -5 - 12 = -17 \] In summary: \[ \begin{array}{lll} \text{i. } \tan \theta & = -\frac{12}{5} \\ \text{ii. } \cos^2 \theta - \sin^2 \theta & = -\frac{119}{169} \\ \text{iii. } 13 \cos \theta + 5 \tan \theta & = -17 \end{array} \]