Suppose that A and B are angles in standard position . Use the given information to find (a) sin(A+B) , (b) sin(A-B) , (c) tan(A+B) , (d) tan(A-B) , (e) the quadrant of A+B and (f) the quadrant of A-B cosA=(12)/(13) and sinB=-(15)/(17),(3pi )/(2)lt Alt 2pi ,(3pi )/(2)lt Blt 2pi (a) sin(A+B)=-(220)/(221) (Simplify your answer,including any radicals . Use integers or fractions for any numbers in the expression.) (b) sin(A-B)=(140)/(221) (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expression.) (c) tan(A+B)=-(220)/(21) (Simplify your answer,including any radicals . Use integers or fractions for any numbers in the expression.) (d) tan(A-B)=square (Simplify your answer,including any radicals . Use integers or fractions for any numbers in the expression.)

1. Determine $sinA$ and $cosB$<br /> Since $cosA = \frac{12}{13}$ and $A$ is in the fourth quadrant, $sinA = -\sqrt{1 - cos^2A} = -\sqrt{1 - \left(\frac{12}{13}\right)^2} = -\frac{5}{13}$. Similarly, since $sinB = -\frac{15}{17}$ and $B$ is in the fourth quadrant, $cosB = \sqrt{1 - sin^2B} = \sqrt{1 - \left(-\frac{15}{17}\right)^2} = \frac{8}{17}$.<br /><br />2. Calculate $sin(A+B)$<br /> Use **$sin(A+B) = sinA \cdot cosB + cosA \cdot sinB$**. Substitute values: $sin(A+B) = \left(-\frac{5}{13}\right)\left(\frac{8}{17}\right) + \left(\frac{12}{13}\right)\left(-\frac{15}{17}\right) = -\frac{40}{221} - \frac{180}{221} = -\frac{220}{221}$.<br /><br />3. Calculate $sin(A-B)$<br /> Use **$sin(A-B) = sinA \cdot cosB - cosA \cdot sinB$**. Substitute values: $sin(A-B) = \left(-\frac{5}{13}\right)\left(\frac{8}{17}\right) - \left(\frac{12}{13}\right)\left(-\frac{15}{17}\right) = -\frac{40}{221} + \frac{180}{221} = \frac{140}{221}$.<br /><br />4. Calculate $tanA$ and $tanB$<br /> Use **$tanA = \frac{sinA}{cosA}$** and **$tanB = \frac{sinB}{cosB}$**. So, $tanA = \frac{-\frac{5}{13}}{\frac{12}{13}} = -\frac{5}{12}$ and $tanB = \frac{-\frac{15}{17}}{\frac{8}{17}} = -\frac{15}{8}$.<br /><br />5. Calculate $tan(A+B)$<br /> Use **$tan(A+B) = \frac{tanA + tanB}{1 - tanA \cdot tanB}$**. Substitute values: $tan(A+B) = \frac{-\frac{5}{12} - \frac{15}{8}}{1 - \left(-\frac{5}{12}\right)\left(-\frac{15}{8}\right)} = \frac{-\frac{5}{12} - \frac{15}{8}}{1 - \frac{75}{96}} = \frac{-\frac{10}{24} - \frac{45}{24}}{\frac{21}{96}} = -\frac{220}{21}$.<br /><br />6. Calculate $tan(A-B)$<br /> Use **$tan(A-B) = \frac{tanA - tanB}{1 + tanA \cdot tanB}$**. Substitute values: $tan(A-B) = \frac{-\frac{5}{12} + \frac{15}{8}}{1 + \left(-\frac{5}{12}\right)\left(-\frac{15}{8}\right)} = \frac{\frac{-10}{24} + \frac{45}{24}}{\frac{21}{96}} = \frac{\frac{35}{24}}{\frac{21}{96}} = \frac{140}{21} = \frac{20}{3}$.<br /><br />7. Determine the quadrants of $A+B$ and $A-B$<br /> Both $A$ and $B$ are in the fourth quadrant, so $A+B$ and $A-B$ will also be in the fourth quadrant.