求证：tan(a+π/4)+tan(a-π/4)=2tan2a

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求证：tan(a+π/4)+tan(a-π/4)=2tan2a
求证：tan(a+π/4)+tan(a-π/4)=2tan2a

求证：tan(a+π/4)+tan(a-π/4)=2tan2a
左边=tan(a+π/4)+tan(a-π/4)
=(tana+1)/(1-tana)+(tana-1)/(1+tana)
=(tana+1)/(1-tana)-(1-tana)/(1+tana)
=[(tana+1)^2-(1-tana)^2]/[1-(tana)^2]
= 2*2tana/[1-(tana)^2]=2tan2a=右边
∴得证

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tan(a+π/4)Xtan(a-π/4)
= (tana+ tanπ/4)/(1-tana tanπ/4)X(tana- tanπ/4)/(1+tana tanπ/4)
=(tana+ 1)/(1-tana )X(tana- 1)/(1+tana )
=1
用tan（A+B）=（tanA+tanB）/（1-tanAtanB)
tan2a =tan(a-π/4+a+π/4)
={tan(a+π/4)+tan(a-π/4)}/{1-tan(a+π/4)Xtan(a-π/4)}
={tan(a+π/4)+tan(a-π/4)}/{1-(-1)}
={tan(a+π/4)+tan(a-π/4)}/2 可得tan(a+π/4)+tan(a-π/4)=2tan2a

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