Question

化简((sinα+cosα-1)(sinα-cosα+1))/sin2α

Answer 1

(sina+cosa-1)(sina-cosa+1)/sin2a

=[sina²-(cosa-1)²]/sin2a

=(1-cos²a-cos²a+2cosa-1)/2sinacosa

=(2cosa-2cos²a)/2sinacosa

=(1-cosa)/sina

=2sin²(a/2)/2sin(a/2)cos(a/2)

=sin(a/2)/cos(a/2)

=tan(a/2)

Answer 2

(sinα+cosα-1)(sinα-cosα+1)/sin2α
= { sinα+(cosα-1)} { sinα-{cosα-1)} / sin2α
= { sin^2α-{cosα-1)^2} / sin2α
= { sin^2α-cos^2α + 2cosα-1 } / sin2α
= { sin^2α-cos^2α + 2cosα-（cos^α+sin^2α) } / sin2α
= { 2cosα-2cos^α } / sin2α
= 2cosα {1-cosα } / (2sinαcosα)
= (1-cosα) / sinα
= tan(α/2)

Answer 3

((sinα+cosα-1)(sinα-cosα+1))/sin2α
={sinα ^2-(cosα-1)^2}/sin2α
={sinα ^2-cosα ^2+2cosα-1}/sin2α
={sinα ^2-cosα ^2+2cosα-sinα ^2-cosα ^2}/sin2α
=2cosα(1-cosα)/(2sinαcosα)
=(1-cosα)/sinα
=tan(α/2)