已知0<x<π/2,化简:lg(cosx·tanx+1–2sin²x/2)+lg[√2cos(x–π/4)]–lg(1+sin2x)
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已知0<x<π/2,化简:lg(cosx·tanx+1–2sin²x/2)+lg[√2cos(x–π/4)]–lg(1+sin2x)
已知0<x<π/2,化简:lg(cosx·tanx+1–2sin²x/2)+lg[√2cos(x–π/4)]–lg(1+sin2x)
已知0<x<π/2,化简:lg(cosx·tanx+1–2sin²x/2)+lg[√2cos(x–π/4)]–lg(1+sin2x)
结果为0