帮下 已知sin(π/4+2a) * sin(π/4-2a) =1/4 ,a∈(π/4,π/2,求2sin^a+tana

sin(π/4+2a) * sin(π/4-2a) =1/4
则(1/2)[cos4a-cos(π/2)]=1/4
cos4a=1/2
a∈(π/4,π/2)
2a∈(π/2,π)
可见cos2a0
由cos4a=2cos²2a-1=1/2 cos²2a=3/4
cos2a=-√3/2 (1)
又cos4a=1-2sin²2a=1/2 sin²2a=1/4
sin2a=1/2 (2)
2sin^a+tana-cota-1
=sina/cosa+cosa/sina-(1-2sin²a)
=(sin²a+cos²a)/(sinacosa)-cos2a
=2/sin2a-cos2a
(1)(2)代入得
原式=2/(1/2)-(-√3/2)
=4+√3/2

sin(π/4+2a) * sin(π/4-2a) =（1/2)[cos4a-cos(π/2)]=1/4,
∴cos4a=1/2,
a∈(π/4,π/2),
4a∈(π,2π),
∴4a=5π/3,a=5π/12,
2(sina)^2=1-cos2a=1-cos(5π/6)=1+(√3）/2，
tana-cota=[(sina)^2-(cosa)^2]/(sinacosa)=-2cos2a/sin2a=-2cot2a=-2cot(5π/6)=2√3，
∴2sin^a+tana-cota-1=（5√3）/2.

sin(π/4+2a) * sin(π/4-2a) =（1/2)[cos4a-cos(π/2)]=1/4,
∴cos4a=1/2,
a∈(π/4,π/2),
4a∈(π,2π),
∴4a=5π/3,a=5π/12,
2(sina)^2=1-cos2a=1-cos(5π/6)=1+(√3）/2，
tana-cota=[(sina)^2-(cosa)^2...

∵sin（π/4+2a)sin(π/4-2a)=[(√2/2)cos2a+(√2/2)sin2a][(√2/2)cos2a-(√2/2)sin2a]=(√2/2)²[cos²2a-sin²2a]=(1/2)cos4a=1/4.
∴cos4a=1/2.又a∈(π/4,π/2),∴4a∈(π，2π）。于是4a=π+π/3=4π/3,即a=π/3或4a=2π-π/3...