已知tanx=2010,则[1-sin(9π/2-2x)]/sin(9π-2x)=

问题描述:

已知tanx=2010,则[1-sin(9π/2-2x)]/sin(9π-2x)=
答案是2010.,但我算出来是1/2010

原式=(1-cos2x)/sin2x
=[1-(1-2sin²x)]/2sinxcosx
=2sin²x/2sinxcosx
=sinx/cosx
=tanx
=2010