Cho $tanB=2$
$A=\dfrac{sinB+cosB}{sinB-cosB}$
= $\dfrac{\dfrac{sinB}{cosB}+\dfrac{cosB}{cosB}}{\dfrac{sinB}{cosB}-\dfrac{cosB}{cosB}}$ (chia cả tử và mẫu cho $cosB$)
= $\dfrac{tanB+1}{tanB-1}$
= $\dfrac{2+1}{2-1}$
= $3$
$B=\dfrac{2sin\alpha+cos\alpha}{3sin\alpha-4cos\alpha}$
= $\dfrac{\dfrac{2sin\alpha}{cos\alpha}+\dfrac{cos\alpha}{cos\alpha}}{\dfrac{3sin\alpha}{cos\alpha}-\dfrac{4cos\alpha}{cos\alpha}}$ (chia cả tử và mẫu cho $cos\alpha$)
= $\dfrac{2tan\alpha+1}{3tan\alpha-4}$
= $\dfrac{2×2+1}{3×2-4}$
= $\dfrac{5}{2}$
