证明:
(sinA+cosA-1)(sinA-cosA+1)/sin2A
=[(sinA)^2-(cosA-1)^2]/sin2A
=((sinA)^2-(cosA)^2+2cosA-1)/sin2A
=2cosA(1-cosA)/2sinAcosA
=(1-cosA)/sinA
=[2(sinA/2)^2]/[2sinA/2 cosA/2]
=(sinA/2)/(cosA/2)
=tanA/2
证明:
(sinA+cosA-1)(sinA-cosA+1)/sin2A
=[(sinA)^2-(cosA-1)^2]/sin2A
=((sinA)^2-(cosA)^2+2cosA-1)/sin2A
=2cosA(1-cosA)/2sinAcosA
=(1-cosA)/sinA
=[2(sinA/2)^2]/[2sinA/2 cosA/2]
=(sinA/2)/(cosA/2)
=tanA/2