[sin(180°)-θsin(270°- θ)tan(90°- θ)]/[sin(90°+ θ)tan(270°- θ)tan(360°- θ)]
题目
[sin(180°)-θsin(270°- θ)tan(90°- θ)]/[sin(90°+ θ)tan(270°- θ)tan(360°- θ)]
答案
[sin(180°-θ)sin(270°- θ)tan(90°- θ)]/[sin(90°+ θ)tan(270°- θ)tan(360°- θ)]
=[sinθ*(-cosθ)costθ]/[cosθ*cotθ*(-tanθ)]
=sinθ/tanθ
=sinθ/(sinθ/cosθ)
=cosθ
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