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

To find the exact value of \( \tan (\alpha - \beta) \), we will use the tangent subtraction formula: \[ \tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \] We already know \( \tan \alpha = -\frac{5}{12} \). Now, we need to find \( \tan \beta \). Since \( \cos \beta = \frac{\sqrt{3}}{2} \) and \( 0 < \beta < \frac{\pi}{2} \), we can find \( \sin \beta \) using the Pythagorean identity: \[ \sin^2 \beta + \cos^2 \beta = 1 \implies \sin^2 \beta + \left(\frac{\sqrt{3}}{2}\right)^2 = 1 \implies \sin^2 \beta + \frac{3}{4} = 1 \] \[ \sin^2 \beta = 1 - \frac{3}{4} = \frac{1}{4} \implies \sin \beta = \frac{1}{2} \] Now we can find \( \tan \beta \): \[ \tan \beta = \frac{\sin \beta}{\cos \beta} = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \] Now, substituting \( \tan \alpha \) and \( \tan \beta \) into the tangent subtraction formula: \[ \tan(\alpha - \beta) = \frac{-\frac{5}{12} - \frac{\sqrt{3}}{3}}{1 + \left(-\frac{5}{12}\right)\left(\frac{\sqrt{3}}{3}\right)} \] To simplify \( -\frac{5}{12} - \frac{\sqrt{3}}{3} \): \[ \frac{\sqrt{3}}{3} = \frac{4\sqrt{3}}{12} \implies -\frac{5}{12} - \frac{\sqrt{3}}{3} = -\frac{5}{12} - \frac{4\sqrt{3}}{12} = -\frac{5 + 4\sqrt{3}}{12} \] Now simplify the denominator \( 1 + \left(-\frac{5}{12}\right)\left(\frac{\sqrt{3}}{3}\right) \): \[ -\frac{5}{12} \cdot \frac{\sqrt{3}}{3} = -\frac{5\sqrt{3}}{36} \implies 1 - \frac{5\sqrt{3}}{36} = \frac{36}{36} - \frac{5\sqrt{3}}{36} = \frac{36 - 5\sqrt{3}}{36} \] Now substitute these back into the formula: \[ \tan(\alpha - \beta) = \frac{-\frac{5 + 4\sqrt{3}}{12}}{\frac{36 - 5\sqrt{3}}{36}} = \frac{- (5 + 4\sqrt{3}) \cdot 36}{12 \cdot (36 - 5\sqrt{3})} = \frac{-3(5 + 4\sqrt{3})}{36 - 5\sqrt{3}} \] Thus, \[ \tan(\alpha - \beta) = \frac{-15 - 12\sqrt{3}}{36 - 5\sqrt{3}} \] To summarize, the final answer is: \[ \boxed{\frac{-15 - 12\sqrt{3}}{36 - 5\sqrt{3}}} \]