(tanα+1)/(1-tanα)=2010 求sec2α+tan2α
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(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
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
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已知(1+tan2α)/(1-tanα)=2010,求(1/cos2α)+tan2α
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