(2cos^2a-1)/(2tan(pai/4+a)sin^2(pai/4-a))
= (cos2a)/{[2sin(π/4+a)/cos(π/4+a)]*sin²(π/4-a)}
∵ sin(π/4+a)=sin[(π/2)-(π/4-a)]=sin(π/4-a)
cos(π/4+a)=cos[(π/2)-(π/4-a)]=sin(π/4-a)
= (cos2a)/{[2cos(π/4-a)/sin(π/4-a)]*sin²(π/4-a)}
=(cos2a)/[2cos(π/4-a)*sin(π/4-a)]
=cos2a/sin(π/2-2a)
=cos2a/(cos2a)
=1