求详解:y=tan(2x+pai/3)-tan(pai/6-2x)的最小正周期求详解,

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求详解:y=tan(2x+pai/3)-tan(pai/6-2x)的最小正周期求详解,

求详解:y=tan(2x+pai/3)-tan(pai/6-2x)的最小正周期求详解,
求详解:y=tan(2x+pai/3)-tan(pai/6-2x)的最小正周期
求详解,

求详解:y=tan(2x+pai/3)-tan(pai/6-2x)的最小正周期求详解,
∵2x+π/3+π/6-2x=π/2
∴y=tan(2x+π/3)-tan(π/6-2x)
=tan(2x+π/3)-cot[π/2-(π/6-2x)]
=tan(2x+π/3)-cot(2x+π/3)
=sin(2x+π/3)/cos(2x+π/3)-cos(2x+π/3)/sin(2x+π/3)
=[sin²(2x+π/3)-cos²(2x+π/3)]/sin(2x+π/3)cos(2x+π/3)
=-2cos2(2x+π/3)/sin2(2x+π/3)
=-2cos(4x+2π/3)/sin(4x+2π/3)
=-2cot(4x+2π/3)
∴y=tan(2x+pai/3)-tan(pai/6-2x)的最小正周期T=π/4