高一数学题(三角函数的诱导公式)若[cos(π+α)*sin^2(-α)]/[sin(π+α)*cos^2(-α)=1/2,则tanα=?

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高一数学题(三角函数的诱导公式)
若[cos(π+α)*sin^2(-α)]/[sin(π+α)*cos^2(-α)=1/2,则tanα=?

这里用a代替α (打字方便)
[cos(π+a)*sin^2(-a)]/[sin(π+a)*cos^2(-a)=1/2
[-cos(a)*sin^2a]/[-sin(a)*cos^2(a)]=1/2
sin(a)/cos(a)=-tan(a)=1/2
tan(a)=1/2

[cos(π+α)*sin^2(-α)]/[sin(π+α)*cos^2(-α)
=[-cos α·sin^2(α)]/[-sin α·cos^2(α)]
=tan α
=1/2

[cos(π+α)*sin^2(-α)]/[sin(π+α)*cos^2(-α)=1/2
[-cosa*sin^2(a)]/[-sina*cos^2(a)]=1/2
sina/cosa=1/2
tana=1/2