Câu `4`:

Ta có:

`(\pi)/2 <\alpha< \pi`

`=> sin \alpha >0; cos \alpha <0`

Ta có:

`tan\alpha . cot \alpha =1`

`<=> tan \alpha = 1/(cot \alpha) = 1/(-3) = -1/3`

Ta có:

`1 + tan^2 \alpha = 1/(cos^2 \alpha)`

`=> cos^2 \alpha . (1 + tan ^2 \alpha) =1`

`<=> cos^2 \alpha =1/(1+ tan^2 \alpha) = 1/(1 + (-1/3)^2 ) =9/10`

`=> cos \alpha = +- (3\sqrt(10))/10`

Do `cos \alpha <0`

`=> cos \alpha = -(3\sqrt(10))/10`

Ta có:

`sin ^2 \alpha + cos^2 \alpha =1`

`=> sin ^2\alpha = 1-cos^2 \alpha`

`=> sin \alpha = +-\sqrt{1-((-3\sqrt(10))/10)^2 }`

Do `sin \alpha >0`

`=> sin \alpha = (\sqrt(10))/10`.