已知(1+tanα)/(1-tanα)=2010,求(1/cos2α)+tan2α的值
问题描述:
已知(1+tanα)/(1-tanα)=2010,求(1/cos2α)+tan2α的值
答
(1+tanα)/(1-tanα)=2010,anα=2009/2011,
(1/cos2α)+tan2α=1/(2cos²α-1)+2anα/(1-tan²α)=1/[2/(1+tan²α)-1]+2anα/(1-tan²α)
=(1+tan²α)/(1-tan²α)+2anα/(1-tan²α)=(1+tanα)²/(1-tan²α)=(1+2009/2011)²/[1-(2009/2011)²]
=2010
或者:
(1/cos2α)+tan2α=1/(2cos²α-1)+2anα/(1-tan²α)=1/[2/(1+tan²α)-1]+2anα/(1-tan²α)
=(1+tan²α)/(1-tan²α)+2anα/(1-tan²α)=(1+tanα)²/(1-tan²α)=(1+tanα)/(1-tanα)=2010anα是什么》。。答:应该是tanα,少打了一个字母t,不好意思,抱歉。