cosα=tanβ，则：
(cosα)^2=(tanβ)^2
=1/(cosβ)^2-1
=1/(tanγ)^2-1
=1/(1/(cosγ)^2-1)-1
=1/(1/(tanα)^2-1)-1
=(tanα)^2/(1-(tanα)^2)-1
=(sinα)^2/((cosα)^2-(sinα)^2)-1
=(1-(cosα)^2)/(2(cosα)^2-1)-1
=(2-3(cosα)^2)/(2(cosα)^2-1)
则：
2(cosα)^4-(cosα)^2=2-3(cosα)^2
2(cosα)^4+2(cosα)^2-2=0
((cosα)^2+1/2)^2=9/4
(cosα)^2+1/2=3/2
(cosα)^2=1
因α为锐角，则(cosα)^2≠1
故本题条件出现矛盾！请重新检查！
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