已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0＜ θ＜派/2.若a·b=0,求tan θ 值

已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0＜ θ＜派/2.若a·b=0,求tan θ 值
已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0＜ θ＜派/2.若a·b=0,求tan θ 值

已知向量a=(sinθ2,cosθ),b=(1,-cos θ ),0＜ θ＜派/2.若a·b=0,求tan θ 值
∵a·b=0 ∴sin^2θ.1+cosθ.-cos θ=0 整理得sin^2θ=cos^2 θ
又∵0＜ θ＜π/2 ∴θ=π/4 故tan θ =1

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