tan(x/2+ π4)+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]
=4tan(x/2)/[1-(tan(x/2))^2]
=2tanx

∵tanπ/4=1
∴tan(x/2+π/4)
=(tanx/2+1)/(1-tanx/2)
tan(x/2-π/4)
=(tana/2-1)/(1+tanx/2)
令a=tanx/2
(a+1)/(1-a)+(a-1)/(1+a)
=[(a+1)^2-(1-a)^2]/(1+a)(1-a)
=2a/(1-a^2)
将a=tanx/2还原：
=2tan(x/2)/[1-tan²(x/2)]
=tan2x