Question

[sin2a/(1+cos2a)]*[cosa/(1+cosa)]*cota/2

Answer 1

原式=[2sinacosa/(1+2cos²a-1)][cosa/(1+cosa)]*cota/2

=(2sinacosa/2cos²a)[cosa/(1+cosa)]*cota/2

=(sina/cosa)[cosa/(1+cosa)]*cota/2

=[sina/(1+cosa)]*cot(a/2)

=[2sin(a/2)cos(a/2)/[1+2cos²(a/2)-1][cos(a/2)/sin(a/2)]

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

=1