`A=sin^2x+2sinx.cosx-3cos^2x`
Có: `1+tan^2x=1/(cos^2x)`
`=> 1/(cos^2x)=1+2^2=5`
`=> cos^2x=1/5`
Lại có:
`A/(cos^2x)=(sin^2x)/(cos^2x)+(2sinx.cosx)/(cos^2x)-(3cos^2x)/(cos^2x)`
$\dfrac{A}{\dfrac{1}{5}}=tan^2x+2.tanx-3$
$\dfrac{A}{\dfrac{1}{5}}=2^2+2.2-3$
$\dfrac{A}{\dfrac{1}{5}}=5$
`A=5. 1/5`
`A=1`
Vậy `A=1` khi `tanx=2`