Question

Suppose A is in quadrant IV and B is in Quadrant IV

Answer 1

From the given data,

`\begin{gathered} cos(A)\text{ = }(15)/(17) \\ sin^2(A)\text{ = 1 -\lparen}(15)/(17))^2 \\ sin^2(A)\text{ = 1 - }(225)/(289) \\ sin^2(A)\text{ =}\frac{289\text{ - 225}}{289} \\ sin^2(A)\text{ = }(64)/(289) \\ sin(A)\text{ = -}(8)/(17)_\text{ \_\_\_\_\_\_\_\lparen Lies on 4th quadrant\rparen} \end{gathered}`

And

`\begin{gathered} cos(B)\text{ = }(9)/(41) \\ sin^2(B)\text{ = 1 - \lparen}(9)/(41))^2 \\ sin^2(B)\text{ = 1 - }(81)/(1681) \\ sin^2(B)\text{ = }\frac{1681\text{ - 81}}{1681} \\ sin^2(B)\text{ = }(1600)/(1681) \\ sin^(B)\text{ = -}(40)/(41)\text{ \_\_\_\_\_\_\_\lparen lies on 4th quadrant\rparen} \\ \end{gathered}`

Calculating the required value,

`\begin{gathered} cos(A-B)\text{ = cosAcosB + sinAsinB} \\ cos(A-B)\text{ = \lparen}(15)/(17))((9)/(41))\text{ + \lparen-}(8)/(17))((-40)/(41)) \\ cos(A-B)\text{ = }(135)/(697)\text{ + }(320)/(697) \\ cos(A-B)\text{ = }(455)/(697) \end{gathered}`

Thus the required answer is,

`cos(A-B)=(455)/(697)`