（tanα+1）/（1-tanα）=2010 求sec2α+tan2α

（tanα+1)/(1-tanα)=(sinα/cosα+1)/(1-sinα/conα)=(sinα+cosα)/(cosα-sinα)=2010
sec(2α)+tan(2α)=1/cos(2α)+sin(2α)/cos(2α)={1+sin(2α)}/cos(2α)
=(sinα+cosα)^2/{(cosα)^2-(sinα)^2}
=(sinα+cosα)/(cosα-sinα)=2010