已知(1+tan2α)/(1-tanα)=2010,求(1/cos2α)+tan2α
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已知(1+tan2α)/(1-tanα)=2010,求(1/cos2α)+tan2α
(1+tan2α)/(1-tanα)=2010
=>{1+2tanα/[(1-tanα)^2]}/(1-tanα)
=1-(tanα)^2+2tanα=2010(1+tanα)
=>2009+(tanα)^2+2008tanα=0 (1)
=>(1+tanα)=[tanα-(tanα)^2]/2009
1/cos2α+tan2α=(1+sin2α)/cos2α=(1+tanα)/(1-tanα)=tanα/2009
根据(1)可求出tanα~=-1,-2007,就可求出上式~=-1/2009 或者-2007/2009.
=>{1+2tanα/[(1-tanα)^2]}/(1-tanα)
=1-(tanα)^2+2tanα=2010(1+tanα)
=>2009+(tanα)^2+2008tanα=0 (1)
=>(1+tanα)=[tanα-(tanα)^2]/2009
1/cos2α+tan2α=(1+sin2α)/cos2α=(1+tanα)/(1-tanα)=tanα/2009
根据(1)可求出tanα~=-1,-2007,就可求出上式~=-1/2009 或者-2007/2009.
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