[2（cosα）^2-1]/2tan(π/4-α)*[sin(π/4+α)]^2化简

[2（cosα）^2-1]/2tan(π/4-α)*[sin(π/4+α)]^2化简
[2（cosα）^2-1]/2tan(π/4-α)*[sin(π/4+α)]^2化简

[2（cosα）^2-1]/2tan(π/4-α)*[sin(π/4+α)]^2化简
注释：此题有可能是书写错误,可能是后面的括号写掉了.
我想原题应该是[2（cosα）^2-1]/[2tan(π/4-α)*[sin(π/4+α)]^2]化简.
它的化简结果是1.解题过程如下：
[2（cosα）^2-1]/[2tan(π/4-α)*[sin(π/4+α)]^2]
=cos（2α）/2*（cos（π/4-α）/sin(π/4-α))*(2/(1-cos(π/2+2α))
=cos（2α）/2*（1+cos（π/2-2α）/sin(π/2-2α))*(2/(1+sin2α))
=(cos（2α）/(1+sin2α))*((1+sin2α)/cos（2α）)
=1