Let `0lt=a , b , c ,dlt=pi,` where `ba n dc` are not complementary, such that `2cosa+6cosb+7cosc+9cosd=0a n d2sina-6sinb+7sinc-9a n dd=0,` then the value of `3(cos(a+d))/("cos"(b+c))` is_____

Answer

7

Explanation

From the given equations, we have
` " " (2 cos a + 9 cos d)^(2) = (6 cos b + 7 cos c)^(2) `
`and ( 2 sin a - 9 sin d)^(2) = (6 sin b - 7 sin c )^(2)`
Adding, we have `36 cos (a+d) = 84 cos (b+c)`
or `(cos(a+d))/(cos (b+c)) = (7)/(3)`