已知等式根号1+sin( θ-π)/1+cos[(π/2)-θ ]=tan(θ+π)-secθ成立,求θ的取值范围
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已知等式根号1+sin( θ-π)/1+cos[(π/2)-θ ]=tan(θ+π)-secθ成立,求θ的取值范围
等号左边全在根号里,
等号左边全在根号里,
![已知等式根号1+sin( θ-π)/1+cos[(π/2)-θ ]=tan(θ+π)-secθ成立,求θ的取值范围](/uploads/image/z/4478773-13-3.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%BC%8F%E6%A0%B9%E5%8F%B71%2Bsin%28+%CE%B8-%CF%80%EF%BC%89%2F1%2Bcos%5B%28%CF%80%2F2%29-%CE%B8+%5D%3Dtan%28%CE%B8%2B%CF%80%EF%BC%89-sec%CE%B8%E6%88%90%E7%AB%8B%2C%E6%B1%82%CE%B8%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
√{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
√[(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
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