Câu `12`

Vì `cos\alpha=3/4 >0` nên `tan\alpha, cot\alpha>0`

Ta có `sin^2 \alpha+ cos^2 \alpha =1`

`=> sin^2 \alpha=1-cos^2 \alpha=1-(3/4)^2=7/16`

`=> sin\alpha=(\sqrt7)/4`

Do đó: `tan \alpha=(sin\alpha)/(cos\alpha)=(\sqrt7)/4 : 3/4=(\sqrt7)/3`

`cot \alpha=1/(tan\alpha)=1: (\sqrt7)/3=(3\sqrt7)/7`

Vậy `B=((\sqrt7)/3 + 3.(3\sqrt7)/7)/((\sqrt7)/3 + (3\sqrt7)/7)=17/8`

Câu `13`

Vì `tan\alpha=\sqrt2>0` nên `cos\alpha>0`

Ta có `1+tan^2\alpha=1/(cos^2\alpha)=> cos^2\alpha=1/(1+tan^2\alpha)=1/(1+2)=1/3`

`=> cos\alpha=(\sqrt3)/3=>cos^2 \alpha=1/3`

`tan\alpha=(sin\alpha)/(cos\alpha)=\sqrt2`

`=> sin\alpha=\sqrt2 . cos\alpha`

Vậy `C=(\sqrt2 . cos\alpha - cos\alpha)/(2\sqrt2 cos^3 \alpha + 3 cos^3 \alpha+2\sqrt2cos\alpha)`

`=((\sqrt2-1)cos\alpha)/(cos\alpha[(2\sqrt2+3)cos^2\alpha+ 2\sqrt2])`

`=((\sqrt2-1))/((2\sqrt2+3)cos^2\alpha+ 2\sqrt2)`

`=((\sqrt2-1))/((2\sqrt2+3). 1/3 + 2\sqrt2)`

`=(\sqrt2-1)/((3+8\sqrt2)/3)`