求证sinx(1+tanx*tan2/x)=tanx如题

用万能公式
sinx=(2sinx/2·cosx/2)/(sin²x/2+cos²x/2)=(2tanx/2)/(1+tan²x/2)
cosx=(cos²x/2-sin²x/2)/(sin²x/2+cos²x/2)=(1-tan²x/2)/(1+tan²x/2)
[sinx(1+tanx*tanx/2)]/tanx=cosx·(1+tanx·tanx/2)
=cosx+cosx·tanx·tanx/2
=cosx+sinx·tanx/2
=(1-tan²x/2)/(1+tan²x/2)+(2tan²x/2)/(1+tan²x/2)
=(1+tan²x/2)/(1+tan²x/2)
=1
∴sinx(1+tanx*tanx/2)=tanx

tanx=2(tanx/2)/[1-(tanx/2)^2]
1+tanx*tanx/2=[1+tan(x/2)^2]/[1-(tanx/2)^2]=[sin(x/2)^2+(cosx/2)^2]/[(cosx/2)^2-(sinx/2)^2]
=1/(-cosx)
sinx*(1+tanx*tanx/2)=sinx*[1/(-cosx)]=-tanx

左边=sinx(1+tanx*tan2/x)
=sinx[1+(sinxsinx/2)/(cosxcosx/2)]
=sinx[sinxsinx/2+cosxcosx/2]/(cosxcosx/2)]
=sinx[cosx/2]/(cosxcosx/2)]
=sinx/cosx
=tanx
=右边
sinx(1+tanx*tan2/x)=tanx

左边化弦
=sinx(cosxcosx/2+sinxsinx/2)/cosxcosx/2
=(sinxcosx/2)/(cosxcosx/2)
=sinx/cosx
=tanx

左边=sinx(1+tanx*tan2/x)
=sinx[1+(sinxsinx/2)/(cosxcosx/2)]
=sinx[sinxsinx/2+cosxcosx/2]/(cosxcosx/2)]
=sinx[cosx/2]/(cosxcosx/2)]
=sinx/cosx
=tanx
=右边
所以sinx(1+tanx*tan2/x)=tanx