tan(x/2+ π4)+tan(x/2- π/4)=2tanx证明
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tan(x/2+π4)+tan(x/2-π/4)=2tanx证明tan(x/2+π4)+tan(x/2-π/4)=2tanx证明tan(x/2+π4)+tan(x/2-π/4)=2tanx证明tan(
tan(x/2+ π4)+tan(x/2- π/4)=2tanx证明
tan(x/2+ π4)+tan(x/2- π/4)=2tanx
证明
tan(x/2+ π4)+tan(x/2- π/4)=2tanx证明
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
tan（ x/2+π/4）+tan（x/2-π/4 ）=2tanxtan(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
这道题为什么第二步的+到了第三步变成了-1.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)-
lim(2-x)tanπ/4x
∫ ( tan^2 x + tan^4 x )dx
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tan(-x+π/4)
(tanπx/4)^(tanπx/2)当x趋向于1时的极限?
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tan(x/2+ π4)+tan(x/2- π/4)=2tanx证明
求证:tan（x/2+π/4）+tan(x/2-π/4)＝2tanx
化简tan(x/2+π/4)-tan（π/4-x/2）
化简 tan(x/2 + π/4)-tan(π/4 - x/2)
解方程：tan(x+π/4)+tan(x-π/4)=2cotx