求证（1+sinθ+cosθ）/（1+sinθ-cosθ）+（1-cosθ+sinθ）/（1+cosθ+sinθ）=2/sinθ如题.请写出求证过程,用高一（下）三角恒等变换知识来证.

原式=(1+sinθ+cosθ)^2/[(1+sinθ)^2-(cosθ)^2]+(1-cosθ+sinθ)^2/[(1+sinθ)^2-(cosθ)^2]=2*(1+sinθ)^2+2(cosθ^2)/[(1+sinθ)^2-(cosθ)^2]= [2+4sinθ+2(sinθ^2)+2(cosθ)^2 ]/ 2sinθ^2 +2sinθ = 2/sinθ
这种证明题关键是在分子分母同时乘以分子对分母进行有理化，剩下的就是化简。

(1＋sinθ＋cosθ)/(1＋sinθ－cosθ)=[2sin(θ/2)cos(θ/2)＋2cos²(θ/2)]/[2sin(θ/2)cos(θ/2)＋2sin²(θ/2)]=cos(θ/2)/sin(θ/2).
(1－cosθ＋sinθ)/(1＋cosθ＋sinθ)=[2sin(θ/2)cos(θ/2)＋sin²(θ/2)]/[2sin(θ/2)cos(θ/2)＋2cos²(θ/2)]=sin(θ/2)/cos(θ/2).
所以,原式=cos(θ/2)/sin(θ/2)＋sin(θ/2)/cos(θ/2)=[cos²(θ/2)＋sin²(θ/2)]/[sin(θ/2)cos(θ/2)]=2/sinθ.