求证tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)=-tanα
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求证tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)=-tanα求证tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-
求证tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)=-tanα
求证tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)=-tanα
求证tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)=-tanα
tan(2π-α)sin(-2π-α)cos(6π-α)/cos(α-π)sin(5π-α)
=(-tana)(-sina)cosa/(-cosa)sina
=-tana
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