tanθ /2=2/3求1—cosθ +sinθ /1+cosθ +sinθ

tan(θ/2)=2/3
由万能公式有
sinθ=2tan(θ/2)/[1+(tan(θ/2))^2]=(4/3)/[1+(2/3)^2]=12/13
cosθ=[1-(tan(θ/2))^2]/[1+(tan(θ/2))^2]=[1-(2/3)^2]/[1+(2/3)^2]=5/13
故(1-cosθ+sinθ)/(1+cosθ +sinθ)
=(1-5/13+12/13)/(1+5/13+12/13)
=2/3