证tan(x/2+兀/4)+tan(x/2-兀/4)=2tanx
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证tan(x/2+兀/4)+tan(x/2-兀/4)=2tanx
证tan(x/2+兀/4)+tan(x/2-兀/4)=2tanx
证tan(x/2+兀/4)+tan(x/2-兀/4)=2tanx
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证tan(x/2+兀/4)+tan(x/2-兀/4)=2tanx
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
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求证 tan(x/2+π/4)+tan(x/2-π/4)=2tan x急急!!!!
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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)-
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求证:tan(x/2+π/4)+tan(x/2-π/4)=2tanx
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