求tan(π-乄)cos(2π-乄)sin(-乄+3π/2)/cos(-乄-π)sin(-π-乄)
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tan(π-x) = tan(x-π) = -tan(π-x)
cos(2π-x) = cos(x)
sin(-x+3π/2) = -cos(x)
cos(-x-π) = -cos(x+π) = -cos(x)
sin(-π-x) = -sin(x)
因此，tan(π-x)cos(2π-x)sin(-x+3π/2)/cos(-x-π)sin(-π-x) = -tan(x)cos(x)(-cos(x))(-sin(x)) = tan(x)sin(x)cos^2(x) = tan(x)(1-sin^2(x)) = tan(x)cos^2(x) = tan(x)(1-tan^2(x))/tan^2(x) = 1/tan(x)
综上，tan(π-x)cos(2π-x)sin(-x+3π/2)/cos(-x-π)sin(-π-x) = 1/tan(x)
cos(2π-x) = cos(x)
sin(-x+3π/2) = -cos(x)
cos(-x-π) = -cos(x+π) = -cos(x)
sin(-π-x) = -sin(x)
因此，tan(π-x)cos(2π-x)sin(-x+3π/2)/cos(-x-π)sin(-π-x) = -tan(x)cos(x)(-cos(x))(-sin(x)) = tan(x)sin(x)cos^2(x) = tan(x)(1-sin^2(x)) = tan(x)cos^2(x) = tan(x)(1-tan^2(x))/tan^2(x) = 1/tan(x)
综上，tan(π-x)cos(2π-x)sin(-x+3π/2)/cos(-x-π)sin(-π-x) = 1/tan(x)
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