tan^2a=2tan^2⊙+1
sin^2a/(1-sin^2a)=2sin^2⊙/(1-sin^2⊙)+1
sin^2a/(1-sin^2a+sin^2a)
=(sin^2⊙+1)/(1-sin^2⊙+sin^2⊙+1)
sin^2a=(sin^2⊙+1)/2
所以：
sin^2⊙=2sin^2a-1 =-cos2a
即cosa+sin^2◎=0
以下是一样的题目,参考一下吧：
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