tan（45+α）=2010=(tan45°+tanα)/(1-tan45°*tan²α)=(1+tanα)/(1-tanα)
解得tanα=2009/2011
cos2α=(cos²α-sin²α)/(cos²α+sin²α)=(1-tan²α)/(1+tan²α)=(1- 2009²/2011²)/(1+ 2009²/2011²)= (2011²-2009²)/(2011²+2009²)
tan2α=2tanα/(1-tan²α)=2*2009*2011/(2011²-2009²)
1/cos2α + tan2α=(2011²+2009²)/(2011²-2009²) + 2*2009*2011/(2011²-2009²)=[(2011²+2009²)+ 2*2009*2011]/(2011²-2009²)=(2011²+2009²+ 2*2009*2011)/(2011²-2009²)=(2011+2009)²/(2011²-2009²)=(2011+2009)²/[(2011+2009)(2011-2009)]=(2011+2009)/(2011-2009)=4020/2=2010