已知tan^2α=2tan^β+1 求证：sin^2β=asin^2α-1快

已知tan^2α=2tan^β+1 求证：sin^2β=asin^2α-1快
已知tan^2α=2tan^β+1 求证：sin^2β=asin^2α-1
快

已知tan^2α=2tan^β+1 求证：sin^2β=asin^2α-1快
题目有误
已知tan^2α=2tan^2β+1 求证：sin^2 β=2sin^2α-1
tan^2α=2tan^2β+1
sin^2α/cos^2α=2sin^2β/cos^2β+1
sin^2α/cos^2α=(2sin^2β+cos^2β)/cos^2β
sin^2α/cos^2α=(sin^2β+1)/cos^2β
sin^2α/(1-sin^2α)=(sin^2β+1)/(1-sin^2β)
sin^2α-sin^2α*sin^2β=sin^2β-sin^2α*sin^2β-sin^2α+1
2sin^2α=sin^2β+1
sin^2β=2sin^2α-1

1楼说的对 你的题目有误！按照1楼的题目，我的证明如下：
tan^2α+1=2(tan^2β+1)
1/cos^2α=2/cos^2β
cos^2β=2cos^2α
1-sin^2β=2(1-sin^2α)
即sin^2 β=2sin^2α-1