已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)

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已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)
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已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)
已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...
已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)

已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x) ...已知0<x<pai/2,化简:lg[cos xtan x+1-2sin^2(pai/2)]+lg[跟号2cos(x-pai/4)]-lg(1+sin2x)
lg[cosxtanx+1-2sin²(x/2)]+lg[√2cos(x-π/4)]-lg(1+sin2x)
=lg(sinx+cosx)+lg(cosx+sinx)-lg[(sinx+cosx) ²]
=lg1=0