(1+tanα)^2+(1+tanα)^2=2/cosα^2拜托各位了 3Q化简:(2)sin^2(α+β)-1 求证:(1+tanα)^2+(1+tanα)^2=2/cosα^2

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(1+tanα)^2+(1+tanα)^2=2/cosα^2拜托各位了 3Q化简:(2)sin^2(α+β)-1 求证:(1+tanα)^2+(1+tanα)^2=2/cosα^2

(1+tanα)^2+(1+tanα)^2=2/cosα^2拜托各位了 3Q化简:(2)sin^2(α+β)-1 求证:(1+tanα)^2+(1+tanα)^2=2/cosα^2
(1+tanα)^2+(1+tanα)^2=2/cosα^2拜托各位了 3Q
化简:(2)sin^2(α+β)-1 求证:(1+tanα)^2+(1+tanα)^2=2/cosα^2

(1+tanα)^2+(1+tanα)^2=2/cosα^2拜托各位了 3Q化简:(2)sin^2(α+β)-1 求证:(1+tanα)^2+(1+tanα)^2=2/cosα^2
(1)sin^2(α+β)-1=-cos^2(a+b)=[-1-cos(2a+2b)]/2证明题目有错吧住呢个出来不等于这个:2(1+tana)^2=2(1+2sina/cosa+sin^2a/cos^2a)=2[(cos^2a+2sinacosa+sin^2a)/cos^2a]=(2+4sinacosa)/cos^2a