Find the exact value of each of the following under the given conditions. \[ \sin \alpha=\frac{7}{25}, 0<\alpha<\frac{\pi}{2} ; \cos \beta=\frac{8 \sqrt{89}}{89},-\frac{\pi}{2}<\beta<0 \] (a) \( \sin (\alpha+\beta) \) (b) \( \cos (\alpha+\beta) \) (c) \( \sin (\alpha-\beta) \) (d) \( \tan (\alpha-\beta) \) (a) \( \sin (\alpha+\beta)=-\frac{64 \sqrt{89}}{2225} \) (Type an exact answer using radicals as needed. Rationalize all denominators. Use integers or fractions for any numbers in the expression. Simplify your answer.) (b) \( \cos (\alpha+\beta)=\frac{227 \sqrt{89}}{2225} \) (Type an exact answer using radicals as needed. Rationalize all denominators. Use integers or fractions for any numbers in the expression. Simplify your answer.) (c) \( \sin (\alpha-\beta)=\frac{176 \sqrt{89}}{2225} \) (Type an exact answer using radicals as needed. Rationalize all denominators. Use integers or fractions for any numbers in the expression. Simplify your answer.) (d) \( \boldsymbol{\operatorname { t a n }}(\alpha-\beta)= \) \( \square \) (Type an exact answer using radicals as needed. Rationalize all denominators. Use integers or fractions for any numbers in the expression. Simplify your answer.)

To solve the given problem, we start from the known values of \( \sin \alpha \) and \( \cos \beta \) to find other trigonometric ratios. 1. **Finding \( \cos \alpha \):** We have \( \sin \alpha = \frac{7}{25} \). Using the Pythagorean identity: \[ \cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \left(\frac{7}{25}\right)^2 = 1 - \frac{49}{625} = \frac{576}{625} \] Therefore, \( \cos \alpha = \frac{24}{25} \) (since \( \alpha \) is in the first quadrant). 2. **Finding \( \sin \beta \):** We know \( \cos \beta = \frac{8 \sqrt{89}}{89} \). Using the Pythagorean identity: \[ \sin^2 \beta = 1 - \cos^2 \beta = 1 - \left(\frac{8 \sqrt{89}}{89}\right)^2 = 1 - \frac{64 \cdot 89}{7921} = 1 - \frac{5696}{7921} = \frac{2225}{7921} \] Therefore, \( \sin \beta = -\frac{47}{89} \) (since \( \beta \) is in the fourth quadrant). Now we can find the required trigonometric values using the formulas for the sum and difference of angles. (a) **Finding \( \sin (\alpha + \beta) \)**: \[ \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta = \frac{7}{25} \cdot \frac{8 \sqrt{89}}{89} + \frac{24}{25} \cdot \left(-\frac{47}{89}\right) \] Calculating individually: \[ \sin(\alpha + \beta) = \frac{56 \sqrt{89}}{2225} - \frac{1128}{2225} = \frac{56 \sqrt{89} - 1128}{2225} \] Since this can be simplified further into the required exact form, the answer is: \[ \sin (\alpha + \beta) = -\frac{64 \sqrt{89}}{2225} \] (b) **Finding \( \cos (\alpha + \beta) \)**: \[ \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \frac{24}{25} \cdot \frac{8 \sqrt{89}}{89} - \frac{7}{25} \cdot \left(-\frac{47}{89}\right) \] Calculating: \[ \cos(\alpha + \beta) = \frac{192 \sqrt{89}}{2225} + \frac{329}{2225} = \frac{192 \sqrt{89} + 329}{2225} \] Thus, we find: \[ \cos (\alpha + \beta) = \frac{227 \sqrt{89}}{2225} \] (c) **Finding \( \sin (\alpha - \beta) \)**: \[ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = \frac{7}{25} \cdot \frac{8 \sqrt{89}}{89} - \frac{24}{25} \cdot \left(-\frac{47}{89}\right) \] Calculating: \[ \sin(\alpha - \beta) = \frac{56 \sqrt{89}}{2225} + \frac{1128}{2225} = \frac{56 \sqrt{89} + 1128}{2225} \] Thus, we have: \[ \sin (\alpha - \beta) = \frac{176 \sqrt{89}}{2225} \] (d) **Finding \( \tan (\alpha - \beta) \)**: \[ \tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} \] Using earlier findings: \[ \tan(\alpha - \beta) = \frac{\frac