Question

（1+sinθ+cosθ）/（1+sinθ-cosθ）=1/2，求tanθ

求各位高手解答！！在线等....这题目我做到不会了- -

Answer 1

(1+sinθ+ cosθ)/(1+sinθ-cosθ)=1/2
2+2sinθ+2cosθ=1+sinθ-cosθ
1+sinθ+3cosθ=0
sinθ=-1-3cosθ ①
两边平方
(sinθ)^2=1+6cosθ+9(cosθ)^2
1-(cosθ)^2=1+6cosθ+9(cosθ)^2
6cosθ+10(cosθ)^2=0
cosθ=0,cosθ=-3/5
若cosθ=0，则sinθ=-1-3cosθ=-1
则分母1+sinθ-cosθ=0，不成立
所以cosθ=-3/5
再根据①得sinθ=4/5
得出tan=-4/3

Answer 2

(1+sinθ+cosθ)/(1+sinθ-cosθ)=1/2,[2sin(θ/2)cos(θ/2)+2cos^2(θ/2)]/[2sin(θ/2)cos(θ/2)+2sin^2(θ/2)]=1/2
cos(θ/2)/sin(θ/2)=1/2
tan(θ/2)=2
tanθ=2tan(θ/2)/[1-tan^2(θ/2)]=4/(1-4)=-4/3