From the given data,
[tex]\begin{gathered} cos(A)\text{ = }\frac{15}{17} \\ sin^2(A)\text{ = 1 -\lparen}\frac{15}{17})^2 \\ sin^2(A)\text{ = 1 - }\frac{225}{289} \\ sin^2(A)\text{ =}\frac{289\text{ - 225}}{289} \\ sin^2(A)\text{ = }\frac{64}{289} \\ sin(A)\text{ = -}\frac{8}{17}_\text{ \_\_\_\_\_\_\_\lparen Lies on 4th quadrant\rparen} \end{gathered}[/tex]
And
[tex]\begin{gathered} cos(B)\text{ = }\frac{9}{41} \\ sin^2(B)\text{ = 1 - \lparen}\frac{9}{41})^2 \\ sin^2(B)\text{ = 1 - }\frac{81}{1681} \\ sin^2(B)\text{ = }\frac{1681\text{ - 81}}{1681} \\ sin^2(B)\text{ = }\frac{1600}{1681} \\ sin^(B)\text{ = -}\frac{40}{41}\text{ \_\_\_\_\_\_\_\lparen lies on 4th quadrant\rparen} \\ \end{gathered}[/tex]
Calculating the required value,
[tex]\begin{gathered} cos(A-B)\text{ = cosAcosB + sinAsinB} \\ cos(A-B)\text{ = \lparen}\frac{15}{17})(\frac{9}{41})\text{ + \lparen-}\frac{8}{17})(\frac{-40}{41}) \\ cos(A-B)\text{ = }\frac{135}{697}\text{ + }\frac{320}{697} \\ cos(A-B)\text{ = }\frac{455}{697} \end{gathered}[/tex]
Thus the required answer is,
[tex]cos(A-B)=\frac{455}{697}[/tex]
