高一数学三角函数题求解

sin^2(α+270°)*tan(α-360°)+sin(-α)*cos(90°+α)*tan^2(α-270°)
=cos^2(a)tana+sin^2(a)*ctan^2(a)
=sinacosa+cos^2(a)
=1/2(2sinacosa+2cos^2(a))
=1/2(sin2a+cos2a-1)
=√2/2(√2/2sin2a+√2/2cos2a)-1/2
=√2/2sin(2a+π/4)-1/2