已知(1+tan2α)/(1-tanα)=2010,求1/cos2α+tan2α的值.
设tana=t,则tan2a=2t/(1-t^2), 已知式变为[1+2t/(1-t^2)]=2010(1-t), 1-t^2+2t=2010(1-t)(1-t^2), 超出中学数学范围,繁! 1/cos2a+tan2a=(1+t^2+2t)/(1-t^2)=(1+t)/(1-t)
答:为方便表达,设tanα=t,则 1/cos2α+tan2α =(1+t^2)/(1-t^2)+2t/(1-t^2) =(1+t)^2/(1-t^2) =(1+t...详情>>
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