Question

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

Answer 1

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

=1/4*tan[2(45-α)]*(sin2α/cos2α)

=1/4*tan(90-2α)]*tan2α

=1/4*cot2αtan2α

=1/4