化简:(1+cos2A)/(ctgA/2-tgA/2)=?
问题描述：
化简:(1+cos2A)/(ctgA/2-tgA/2)=?
答
1+cos2A
= 1+2cos²A-1
= 2cos²A
ctg(A/2)-tg(A/2)
=cos(A/2)/sin(A/2)-sin(A/2)/cos(A/2)
=cos²(A/2)/[cos(A/2)sin(A/2)]-sin²(A/2)/[cos(A/2)sin(A/2)]
=[cos²(A/2)-sin²(A/2)]/[cos(A/2)sin(A/2)]
=cosA/[(1/2)sinA]
=2cosA/sinA
所以
(1+cos2A)/(ctgA/2-tgA/2)
= 2cos²A/【2cosA/sinA】
= 2cos²AsinA/2cosA
=cosAsinA
=（1/2）sin（2A）