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

To find \(\cos(\alpha + \beta)\), we use the formula: \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \] First, we need to find \(\sin \alpha\) and \(\cos \alpha\). From the given \(\tan \alpha = -\frac{5}{12}\) and noting that \(\alpha\) is in the second quadrant: - Opposite side = -5 (since tangent is negative in the second quadrant) - Adjacent side = 12 (positive in the second quadrant) Using the Pythagorean theorem: \[ \sqrt{(-5)^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \] Thus, we have: - \(\sin \alpha = \frac{opposite}{hypotenuse} = \frac{-5}{13} \quad (\text{negative in quadrant II})\) - \(\cos \alpha = \frac{adjacent}{hypotenuse} = \frac{12}{13} \quad (\text{positive in quadrant II})\) For \(\beta\), since \(\cos \beta = \frac{\sqrt{3}}{2}\) and \(\beta\) is in the first quadrant, we have: - \(\sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\) Now substituting into our formula: \[ \cos(\alpha + \beta) = \left(\frac{12}{13}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{-5}{13}\right) \left(\frac{1}{2}\right) \] Calculating this gives: \[ \cos(\alpha + \beta) = \frac{12\sqrt{3}}{26} + \frac{5}{26} = \frac{12\sqrt{3} + 5}{26} \] So the answer for part (b) is: (b) \( \cos (\alpha+\beta) = \frac{12\sqrt{3} + 5}{26} \)