证明:1+tan^2a=tana/sinacosa

证明 (sinAcosA)/(cos^2A-sin^2A)=tanA/(1-tan^2A) 证明2tana/1-tan^2a证明2tana/1-tan^2a=tan2a 已知tan(a-r)/tana+sin^2b/sin^2a=1 ,求证tan^2b=tana*tanr.. 已知tan(A-B)/tanA+sin^2C/sin^2A=1,求证：tanA*tanB=tan^2C 已知tan(A-B)/tanA+sin^2C/sin^2A=1,求证：tanA×tanB＝tan^C求证tanA×tanB＝tan^2C tan(a+b)=3tana 证明2sin2b-sin2a=sin(2a+2b) 已知3Sin2b=sin(2a+b)证明:tan(a+b)=2tana 证明:1+tan^2a=tana/sinacosa 不用万能公式,证明tanA=(2tanA/2)/(1-tan^2(A/2)) 证明下列各式:(1+tana+cota)/(1+tan^2 a+tana)-cota/(1+tan^2 a)=sinacosa 证明,tan（A/2）-1/tan(A/2)=-2/tana tan(A/x)-1/tan(A/2)=-2/tanA这个怎么证明呢 证明tan3a=3tana-tan^3a/1-3tan^2a 三角函数证明证明下列恒等式：（1）tan^2a×sin^2a=tan^2a-sin^2a (2)cosa+tana--------------- =sinacosa/sina+1/cosa(3) cos^2a-sin^2a 1-tana ----------------------- =----------------1+2sinacosa 1+tana 知tan(A-B)/tanA+sin^2C/sin^2A=1求证tanAtanB=tan^2C 证明(1-2sinAcosA)/(cos^2A-sin^2A)=(1-tanA)/(1+tanA) cos^2a-sin^2a/1+2sinacosa=1-tana/1+tana怎么证明? 证明下列恒等式1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α证明下列恒等式:1、cos^2α+2sin^2α+sin^2αtan^2α=1/cos^2α2、cos^2α(2+tanα)(1+2tanα)=2+5sinαcosα3、(1+tan^2A)/(1+cot^2A)=[(1-tanA)/(1-cotA)]^24、(tanA-tanB)/(cotB-cotA