Cho `tan\alpha + cot\alpha = 3` Tính `A = sin\alpha . cos\alpha`

`tan alpha + cot alpha = 3`

`=> [sin alpha]/[cos alpha]+[cos alpha]/[sin alpha]=3`

`=> [sin^2 alpha]/[cos alpha . sin alpha]+[cos^2 alpha]/[cos alpha . sin alpha]=3`

`=> [sin^2 alpha+cos^2 alpha]/[cos alpha . sin alpha]=3`

`=> 1/[cos alpha . sin alpha]=3`

`=> cos alpha . sin alpha = 1/3`

Vậy `A=cos alpha . sin alpha =1/3`