Find the exact value of each of the following under the given conditions below. \[ \begin{array}{llll}\tan \alpha=-\frac{12}{5}, \frac{\pi}{2}<\alpha<\pi ; \sin \beta=\frac{\sqrt{3}}{2}, 0<\beta<\frac{\pi}{2} \\ \begin{array}{llll}\text { (a) } \sin (\alpha+\beta) & \text { (b) } \cos (\alpha+\beta) & \text { (c) } \sin (\alpha-\beta) & \text { (d) } \tan (\alpha-\beta)\end{array}\end{array} . \]

To find the exact values of \(\sin(\alpha + \beta)\), \(\cos(\alpha + \beta)\), \(\sin(\alpha - \beta)\), and \(\tan(\alpha - \beta)\), we will use the given values and the angle sum and difference formulas, along with the definitions of trigonometric functions. ### Step 1: Identify values for \(\alpha\) and \(\beta\) We know: - \(\tan \alpha = -\frac{12}{5}\) and \(\frac{\pi}{2} < \alpha < \pi\). - \(\sin \beta = \frac{\sqrt{3}}{2}\) and \(0 < \beta < \frac{\pi}{2\). From \(\sin \beta = \frac{\sqrt{3}}{2}\), it follows that: \[ \beta = \frac{\pi}{3} \] ### Step 2: Determine \(\cos \alpha\) and \(\sin \alpha\) Using \(\tan \alpha = -\frac{12}{5}\), we write: \[ \sin \alpha = -12k, \quad \cos \alpha = 5k \] because \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\). Using the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] gives us: \[ (-12k)^2 + (5k)^2 = 1 \implies 144k^2 + 25k^2 = 1 \implies 169k^2 = 1 \implies k^2 = \frac{1}{169} \implies k = \frac{1}{13} \] Thus: \[ \sin \alpha = -\frac{12}{13}, \quad \cos \alpha = \frac{5}{13} \] ### Step 3: Calculate \(\sin(\alpha + \beta)\) Using the formula: \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \] where \(\cos \beta = \frac{1}{2}\) (from \(\beta = \frac{\pi}{3}\)): \[ \sin(\alpha + \beta) = \left(-\frac{12}{13}\right) \left(\frac{1}{2}\right) + \left(\frac{5}{13}\right) \left(\frac{\sqrt{3}}{2}\right) = -\frac{6}{13} + \frac{5\sqrt{3}}{26} = -\frac{12}{26} + \frac{5\sqrt{3}}{26} = \frac{5\sqrt{3} - 12}{26} \] ### Step 4: Calculate \(\cos(\alpha + \beta)\) Using the formula: \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]: \[ \cos(\alpha + \beta) = \left(\frac{5}{13}\right) \left(\frac{1}{2}\right) - \left(-\frac{12}{13}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{5}{26} + \frac{12\sqrt{3}}{26} = \frac{5 + 12\sqrt{3}}{26} \] ### Step 5: Calculate \(\sin(\alpha - \beta)\) Using the formula: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \]: \[ \sin(\alpha - \beta) = \left(-\frac{12}{13}\right) \left(\frac{1}{2}\right) - \left(\frac{5}{13}\right) \left(\frac{\sqrt{3}}{2}\right) = -\frac{6}{13} - \frac{5\sqrt{3}}{26} = -\frac{12}{26} - \frac{5\sqrt{3}}{26} = -\frac{12 + 5\sqrt{3}}{26} \] ### Step 6: Calculate \(\tan(\alpha - \beta)\) Using the formula: \[ \tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} \] where \(\cos(\alpha