证明,tanatan2a/tan2a-tana=sin2a=2{sina/cosa}/cos^2a/cos^2+sin^2/cos^2分子分母同时成cos^2= [2tanα·(cosα)^2]/[(cosα)^2 + (sinα)^2]= (2sinαcosα)/1= sin2α

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证明,tanatan2a/tan2a-tana=sin2a=2{sina/cosa}/cos^2a/cos^2+sin^2/cos^2分子分母同时成cos^2=[2tanα·(cosα)^2]/[(c

证明,tanatan2a/tan2a-tana=sin2a=2{sina/cosa}/cos^2a/cos^2+sin^2/cos^2分子分母同时成cos^2= [2tanα·(cosα)^2]/[(cosα)^2 + (sinα)^2]= (2sinαcosα)/1= sin2α
证明,tanatan2a/tan2a-tana=sin2a
=2{sina/cosa}/cos^2a/cos^2+sin^2/cos^2
分子分母同时成cos^2
= [2tanα·(cosα)^2]/[(cosα)^2 + (sinα)^2]
= (2sinαcosα)/1
= sin2α

证明,tanatan2a/tan2a-tana=sin2a=2{sina/cosa}/cos^2a/cos^2+sin^2/cos^2分子分母同时成cos^2= [2tanα·(cosα)^2]/[(cosα)^2 + (sinα)^2]= (2sinαcosα)/1= sin2α
分子 = tanα·tan2α = tanα·2tanα/[1 - (tanα)^2] ,分母 = tan2α - tanα = 2tanα/[1 - (tanα)^2] - tanα ,分子分母同时乘以 [1 - (tanα)^2] ,
原式 = tanα·2tanα/{2tanα - tanα·[1 - (tanα)^2]}
= 2(tanα)^2/[tanα + (tanα)^3]
= 2tanα/[1 + (tanα)^2]
= [2tanα·(cosα)^2]/[(cosα)^2 + (sinα)^2]
= (2sinαcosα)/1
= sin2α
证毕.

求证:tan3a-tan2a-tana=tan3a-tanatan2a tanatan2a/(tan2a-tana)+根号3(sin^2a-cos^2)=2sin(2a-派/3) 证明下sin2a怎么来得请说详细 tanαtan2α/(tan2α+tanα)=sin2α证明tanatan2a/(tan2a-tana)+根号3(sin^2a-co^2)=2sin(2a-派/3) 证明下化简tanatan2a/tan2a-tana 的结果A sina B cosa C sin2a D cos2a 证明 tan2A - sin2A=tan2A*sin2A 证明,tanatan2a/tan2a-tana=sin2a=2{sina/cosa}/cos^2a/cos^2+sin^2/cos^2分子分母同时成cos^2= [2tanα·(cosα)^2]/[(cosα)^2 + (sinα)^2]= (2sinαcosα)/1= sin2α 急求三角函数几个公式证明过程及角A取值范围设tanA=tsin2A=2t/(1+t^2) tan2A=2t/(1-t^2) cos2A=(1-t^2)/(1+t^2) tan2A-tanA/tan2A+cotA=tan^2A 请问如何证明, 数学正切证明题tan5a-tan3a-tan2a=tan5a*tan3a*tan2a 证明:tan2a= 2tana/(1-tana^2) cos2A=1/(tan2A+1)对吗?证明2为平方证明(tan2atana)/(tan2a-tana)=sin2a如题证明2tana/1-tan^2a证明2tana/1-tan^2a=tan2a 证明tan(π/4+a)-cot(π/4+a)=2tan2a tan2a=2tana/1-tan平方a=4/3 怎样用初中的知识点证明! 急 ,sin2A/sin2B+cos2Acos2C=1,证明tan2A/tan2B=sin2C 证明恒等式(tan2a-sin2a)cot2a=sin2a.2指的是平方用数学归纳法证明tana·tan2a+tan2a·tan3a+…+tan(n-1)a·tanna=tanna/tana-n(n≥2,n∈N+)