若tanx/2=2/3,化简1-cosx+sinx/1+cosx+sinx

问题描述:

若tanx/2=2/3,化简1-cosx+sinx/1+cosx+sinx

tanx/2=2/3
(1-cosx+sinx)/(1+cosx+sinx)
=[(cosx/2+sinx/2)^2-(cos^2x/2-sin^2x/2)]/[(cosx/2+sinx/2)^2+(cos^2x/2-sin^2x/2)]
=2(cosx/2+sinx/2)*sinx/2/2(cosx/2+sinx/2)*cosx/2
=sinx/2/cosx/2
=tanx/2
=2/3