Đáp án:

Ta có : `sin^2 alpha + cos^2 alpha = 1`

`=> sin^2 alpha + (2/3)^2 = 1`

`=> sin^2 alpha + 4/9 = 1`

`=> sin^2 alpha = 5/9`

`=> sin alpha = (sqrt{5})/3` hoặc `sin alpha = -(sqrt{5})/3`

Với `sin alpha = (sqrt{5})/3 => tan alpha = (sin alpha)/(cos alpha) = (sqrt{5})/2`

`cot alpha = 1/(tan alpha) = (2sqrt{5})/5`

`=> A = ( (sqrt{5})/2 + 3 . (2sqrt{5})/5)/( (sqrt{5})/2 + (2sqrt{5})/5)`

         ` = 17/9`

Với `sin alpha = -(sqrt{5})/3 => tan alpha = (sin alpha)/(cos alpha) = -(sqrt{5})/2`

`cot alpha = 1/(tan alpha) =-(2sqrt{5})/5`

`=> A = ( -(sqrt{5})/2 - 3 . (2sqrt{5})/5)/( -(sqrt{5})/2 - (2sqrt{5})/5)`

        ` = 17/9`

Vậy `A = 17/9`