tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanx
问题描述：
tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanx
tan(x/2+π/4)+tan(x/2-π/4)
=[tan(x/2)+tan(π/4)]/[1-tan(x/2)tan(π/4)]+[tan(x/2)-tan(π/4)]/[1+tan(x/2)tan(π/4)]
=[tan(x/2)+1]/[1-tan(x/2)]+[tan(x/2)-1]/[1+tan(x/2)]
=[(tan(x/2)+1)^2-(tan(x/2)-1)^2]/[1-(tan(x/2))^2]
这里开始看不懂了.=4tan(x/2)/[1-(tan(x/2))^2]
=2tanx
答
分子把平方展开之后整个式子化为4tan(x/2)/[1-(tan(x/2))^2]=2{tan(x/2)+tan(x/2)/[1-(tan(x/2))×(tan(x/2))]}=2tanx。。=4tan(x/2)/[1-(tan(x/2))^2]是这部不懂啊把分子的平方都展开