设a=A则
(1/tanA/2-tanA/2)(1+tanAtanA/2)
=[(1-tan²A/2)/tanA/2][1+2tan²A/2/(1-tan²A/2)]
=[(1-tan²A/2)/tanA/2][(1+tan²A/2)/(1-tan²A/2)]
=(1+tan²A/2)/tanA/2
=sec²A/2/tanA/2
=2/sinA
设a=A则
(1/tanA/2-tanA/2)(1+tanAtanA/2)
=[(1-tan²A/2)/tanA/2][1+2tan²A/2/(1-tan²A/2)]
=[(1-tan²A/2)/tanA/2][(1+tan²A/2)/(1-tan²A/2)]
=(1+tan²A/2)/tanA/2
=sec²A/2/tanA/2
=2/sinA