若(1+tanα)/(1-tanα)=2009,则 1/(cos2α)+tan2α=___.〖参考答案〗2009我只有答案.请尽量写出解析过程.
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若(1+tanα)/(1-tanα)=2009,则 1/(cos2α)+tan2α=___.〖参考答案〗2009我只有答案.请尽量写出解析过程.
若(1+tanα)/(1-tanα)=2009,则 1/(cos2α)+tan2α=___.
〖参考答案〗2009
我只有答案.
请尽量写出解析过程.
若(1+tanα)/(1-tanα)=2009,则 1/(cos2α)+tan2α=___.〖参考答案〗2009我只有答案.请尽量写出解析过程.
1/(cos2α)+tan2α
=1/(cos²α-sin²α)+tan2α
=(sin²α+cos²α)/(cos²α-sin²α)+tan2α
=(tan²α+1)/(1-tan²α)+2tanα/(1-tan²α)
=(tanα+1)²/(1+tanα)(1-tanα)
=(1+tanα)/(1-tanα)=2009
答案应该是2008
(1+tanα)/(1-tanα)=2008
上下同乘以cosα后平方
得(1+sinα)/(1-sinα)
=2008²(1/cos2α)+tan2α
=(1+sin2α)/cosα
两边平方原式²=(1+sin2α)²/(1-sin²2α)
=(1+sinα)/(1-sinα)
=2008²所以原式=2008
1/(cos2α)+tan2α=2010