Find the exact value of each of the following under the given conditions below. \[ \begin{array}{lll}\tan \alpha=-\frac{5}{12}, \frac{\pi}{2}<\alpha<\pi ; \cos \beta=\frac{\sqrt{3}}{2}, 0<\beta<\frac{\pi}{2} \\ \begin{array}{lll}\text { (a) } \sin (\alpha+\beta) & \text { (b) } \cos (\alpha+\beta) & \text { (c) } \sin (\alpha-\beta)\end{array} \quad \text { (d) } \tan (\alpha-\beta) \\ \text { (a) } \sin (\alpha+\beta)=\frac{5 \sqrt{3}-12}{26} \\ \text { (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) } \\ \text { (b) } \cos (\alpha+\beta)=\frac{-12 \sqrt{3}-5}{26} & \text { (Simplify your answer, ineluding any radicals. Use integers or fractions for any numbers in the expression.) } \\ \text { (c) sin }(\alpha-\beta)=\square & \text { (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) }\end{array} \]

To find the values of \(\sin(\alpha-\beta)\) and \(\tan(\alpha-\beta)\), we first need to determine the sine and cosine of \(\alpha\) based on the given \(\tan \alpha = -\frac{5}{12}\). Since \(\alpha\) is in the second quadrant (\(\frac{\pi}{2}<\alpha<\pi\)), we can use the Pythagorean identity to find \(\sin \alpha\) and \(\cos \alpha\). Using \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\), we have: \[ \sin \alpha = -5k \quad \text{and} \quad \cos \alpha = 12k \] for some \(k\). Using the identity \(\sin^2 \alpha + \cos^2 \alpha = 1\): \[ (-5k)^2 + (12k)^2 = 1 \implies 25k^2 + 144k^2 = 1 \implies 169k^2 = 1 \implies k^2 = \frac{1}{169} \implies k = \frac{1}{13} \] Thus: \[ \sin \alpha = -\frac{5}{13} \quad \text{(negative in the second quadrant)} \quad \text{and} \quad \cos \alpha = -\frac{12}{13} \quad \text{(negative in the second quadrant)}. \] Next, for \(\beta\): Since \(\cos \beta = \frac{\sqrt{3}}{2}\) and \(0 < \beta < \frac{\pi}{2}\), we know: \[ \sin \beta = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \] Now, we can calculate \(\sin(\alpha - \beta)\): \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] Substituting the known values: \[ \sin(\alpha - \beta) = \left(-\frac{5}{13}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(-\frac{12}{13}\right)\left(\frac{1}{2}\right) \] \[ = -\frac{5\sqrt{3}}{26} + \frac{12}{26} = \frac{12 - 5\sqrt{3}}{26} \] Now, for \(\tan(\alpha - \beta)\): \[ \tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} \] Calculating \(\cos(\alpha - \beta)\): \[ \cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \] Substituting: \[ = \left(-\frac{12}{13}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{5}{13}\right)\left(\frac{1}{2}\right) \] \[ = -\frac{12\sqrt{3}}{26} - \frac{5}{26} = -\frac{12\sqrt{3} + 5}{26} \] Finally, combining everything: \[ \tan(\alpha - \beta) = \frac{\frac{12 - 5\sqrt{3}}{26}}{-\frac{12\sqrt{3} + 5}{26}} = \frac{12 - 5\sqrt{3}}{-(12\sqrt{3} + 5)} = -\frac{12 - 5\sqrt{3}}{12\sqrt{3} + 5} \] So the final answers are: \[ \text{(c) } \sin(\alpha - \beta) = \frac{12 - 5\sqrt{3}}{26} \] \[ \text{(d) } \tan(\alpha - \beta) = -\frac{12 - 5\sqrt{3}}{12\sqrt{3} + 5}