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(1)设x=sqrt(3)/2*tany-1/21-fn(x)=(sqrt(3)*tany-n)/(1/2*(secy)^2)=2sqrt(3)siny*cosy-2n*(cosy)^2==sqrt(3)sin(2y)-n(1+cos(2y))=sqrt(3+n^2)sin(2y+z)-ncn=(1-fn(x))min*(1-fn(x))max=(-sqrt(3+n^2)-n)(sqrt(3+n^2)-n)=n^2-(3+n^2)=-3故数列{cn}是常数项选C，是常数列(2)延长AP,BP,CP,DP到A',B',C',D'使得AP=A'P.BP=B'P,CP=C'P.DP=D'P易证ABCD-C'D'A'B' 是... |
