已知sin平方2a+sin2acosa-cos2a=1,a属于(0,90),求sina,tana

∵sin²(2a)+sin(2a)cosa-cos(2a)=1
∴sin²(2a)+sin(2a)cosa-cos(2a)-1=0
==>sin²(2a)+sin(2a)cosa-2cos²a=0
==>[sin(2a)+2cosa][sin(2a)-cosa]=0
==>(2sinacosa+2cosa)(2sinacosa-cosa)=0
==>2cos²a(sina+1)(2sina-1)=0
∵a属于(0,90)
∴cosa>0 ==>cos²a≠0
sina>0 ==>sina+1≠0
∴2sina-1=0 ==>sina=1/2
==>a=π/6
故sina=1/2
tana=√3/3

(sin2a)平方+sin2acosa-cos2a=1
4sina^2cosa^2+2sinacosa^2-2cosa^2+1=1
4sina^2cosa^2+2sinacosa^2-2cosa^2=0
由于a的范围，所以cosa^2不等于0
则4sina^2+2sina-2=0
sina=-1或1/2
又a属于（0，π/2），所以a=π/6
所以sina=1/2,tana=√3/3