【高中数学】△ABC中,∠BCA=90°,AB=1,过C作CD⊥AB于D,过点A△ABC中,∠BCA=90°,AB=1,过C作CD⊥AB于D,过点A作AE⊥AC,CD的延长线交AE于E,设∠B=θ,θ是变量.(1)求证:CD-DE=tanθ•cos2θ.(2)记y=6/5(CA+CB)-CD,
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![【高中数学】△ABC中,∠BCA=90°,AB=1,过C作CD⊥AB于D,过点A△ABC中,∠BCA=90°,AB=1,过C作CD⊥AB于D,过点A作AE⊥AC,CD的延长线交AE于E,设∠B=θ,θ是变量.(1)求证:CD-DE=tanθ•cos2θ.(2)记y=6/5(CA+CB)-CD,](/uploads/image/z/4027141-37-1.jpg?t=%E3%80%90%E9%AB%98%E4%B8%AD%E6%95%B0%E5%AD%A6%E3%80%91%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0BCA%3D90%C2%B0%2CAB%3D1%2C%E8%BF%87C%E4%BD%9CCD%E2%8A%A5AB%E4%BA%8ED%2C%E8%BF%87%E7%82%B9A%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0BCA%3D90%C2%B0%2CAB%3D1%2C%E8%BF%87C%E4%BD%9CCD%E2%8A%A5AB%E4%BA%8ED%2C%E8%BF%87%E7%82%B9A%E4%BD%9CAE%E2%8A%A5AC%2CCD%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%A4AE%E4%BA%8EE%2C%E8%AE%BE%E2%88%A0B%3D%CE%B8%2C%CE%B8%E6%98%AF%E5%8F%98%E9%87%8F.%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9ACD-DE%3Dtan%CE%B8%26%238226%3Bcos2%CE%B8.%EF%BC%882%EF%BC%89%E8%AE%B0y%3D6%2F5%28CA%2BCB%29-CD%2C)
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