（1+tanα）/（1-tanα）=2010,anα=2009/2011,
(1/cos2α)+tan2α=1/(2cos²α-1)+2anα/(1-tan²α)=1/［2/(1+tan²α)-1］+2anα/(1-tan²α)
=(1+tan²α)/(1-tan²α)+2anα/(1-tan²α)=(1+tanα)²/(1-tan²α)=(1+2009/2011)²/［1-(2009/2011)²］
=2010
或者：
(1/cos2α)+tan2α=1/(2cos²α-1)+2anα/(1-tan²α)=1/［2/(1+tan²α)-1］+2anα/(1-tan²α)
=(1+tan²α)/(1-tan²α)+2anα/(1-tan²α)=(1+tanα)²/(1-tan²α)=(1+tanα)/(1-tanα)=2010anα是什么》。。答：应该是tanα，少打了一个字母t，不好意思，抱歉。