Question

2tan(45°-α)]/[1-tan²(45°-α)]*[（sinα*cosα）/（cos²α-sin²α）]

Answer 1

2tan(45°-α)]/[1-tan²(45°-α)]*[（sinα*cosα）/（cos²α-sin²α）]
=tan2(45°-α)X1/2xtan2α
=tan(90°-2α)x1/2xtan2α
=cot2αtan2αx1/2
=1/2

Answer 2

2tan(45°-α)]/[1-tan²(45°-α)]*[（sinα*cosα）/（cos²α-sin²α）]
=tan[2(45°-α)]*[（sinα*cosα）/（1-2sin²α）]
=cot2α*（sinα*cosα）/cos2α
=cos2α/sin2α*1/2 sin2α/cos2α
=1/2

Answer 3

2tan(45°-α)]/[1-tan²(45°-α)]*[（sinα*cosα）/（cos²α-sin²α）]
=tan[2(45°-α)]（1/2*sin2α）/（cos2α）
=cot2α（1/2*sin2α）/（cos2α）
=1/2