sin(-19π/6)/(sec4π/3)+tan7π/4=
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sin(-19π/6)/(sec4π/3)+tan7π/4=sin(-19π/6)/(sec4π/3)+tan7π/4=sin(-19π/6)/(sec4π/3)+tan7π/4=原式=sin(-19
sin(-19π/6)/(sec4π/3)+tan7π/4=
sin(-19π/6)/(sec4π/3)+tan7π/4=
sin(-19π/6)/(sec4π/3)+tan7π/4=
原式=sin(-19π/6+4π)cos(π+π/3)+tan(π+3/4π)
=-sin(5π/6)cos(π/3)+tan(3π/4)
=-1/2×1/2-1
=-5/4
sin(-19π/6)/(sec4π/3)+tan7π/4=
sin(-19π/6)/(sec4π/3)+tan7π/4=详细过程谢谢这是道选择题A1/6B-1/6C√3/3D-√3/3
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