`a. cos x =2/5 (-pi/2<x<0)`

Vì `(-pi/2<x<0)` nên `sin x <0`

Ta có : `sin ^2 x +cos^2 x =1`

`=> sin^2 x =1-cos^2 x`

`=> sin^2 x=1-(2/5)^2`

`=>sin^2 x=21/25`

Vì `sin x<0` `=> sin x=-sqrt21/5` 

`b. sin alpha =1/3`

`A=(cot alpha -tan alpha)/(tan alpha +2 cot alpha)`

`A=[cos alpha/sin alpha-sin alpha/cos alpha]/[sin alpha /cos alpha+2cos alpha/sin alpha]`

`A=[(cos^2 alpha-sin ^2alpha)/(sin alpha .cos alpha)]/[(sin ^2 alpha + 2 cos^2 alpha)/(cos alpha.sin alpha)`

`A=[cos^2 alpha-sin^2 alpha]/[sin^2 alpha+2cos^2 alpha]`

`A= (1-2 sin ^2 alpha)/(sin^2 alpha +2.(1-sin^2 alpha)`

`A= (1-2.(1/3)^2)/((1/3)^2+2.[1-(1/3)^2]`

`A=7/17`