√{1+sin( θ-π）/1+cos[(π/2)-θ ]}=tan(θ+π）-secθ
√[(1-sinθ)/(1+sinθ)]=tanθ-1/cosθ
(cosθ/2-sinθ/2)/(cosθ/2+sinθ/2)=(sinθ-1)/(cosθ/2-sinθ/2)(cosθ/2+sinθ/2)
(cosθ/2-sinθ/2)^2=sinθ-1
1-sinθ=sinθ-1
sinθ=1
θ=2kπ+π/2