作业帮已知(sinα+cosα)/(sinα-cosα)=3,求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²高手帮帮忙啊~~~已知(sinα+cosα)/(sinα-cosα)=3，求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²α)]不好意思
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/13 06:40:18
![已知(sinα+cosα)/(sinα-cosα)=3,求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²高手帮帮忙啊~~~已知(sinα+cosα)/(sinα-cosα)=3，求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²α)]不好意思](/uploads/image/z/1105037-53-7.jpg?t=%E5%B7%B2%E7%9F%A5%28sin%CE%B1%2Bcos%CE%B1%29%2F%28sin%CE%B1-cos%CE%B1%29%3D3%2C%E6%B1%82%5Bsin%26%23178%3B%EF%BC%882%CF%80-%CE%B1%EF%BC%89%2Bcot%EF%BC%883%2F2%2A%CF%80-%CE%B1%EF%BC%89%5D%2F%5Btan%28%CF%80%2B%CE%B1%29%281%2Bsin%26%23178%3B%E9%AB%98%E6%89%8B%E5%B8%AE%E5%B8%AE%E5%BF%99%E5%95%8A%7E%7E%7E%E5%B7%B2%E7%9F%A5%28sin%CE%B1%2Bcos%CE%B1%29%2F%28sin%CE%B1-cos%CE%B1%29%3D3%EF%BC%8C%E6%B1%82%5Bsin%26%23178%3B%EF%BC%882%CF%80-%CE%B1%EF%BC%89%2Bcot%EF%BC%883%2F2%2A%CF%80-%CE%B1%EF%BC%89%5D%2F%5Btan%28%CF%80%2B%CE%B1%29%281%2Bsin%26%23178%3B%CE%B1%29%5D%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D)
已知(sinα+cosα)/(sinα-cosα)=3,求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²高手帮帮忙啊~~~已知(sinα+cosα)/(sinα-cosα)=3，求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²α)]不好意思
已知(sinα+cosα)/(sinα-cosα)=3,求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²
高手帮帮忙啊~~~
已知(sinα+cosα)/(sinα-cosα)=3，求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²α)]
不好意思啊，我以为全发了的说……
已知(sinα+cosα)/(sinα-cosα)=3,求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²高手帮帮忙啊~~~已知(sinα+cosα)/(sinα-cosα)=3，求[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²α)]不好意思
将(sinα+cosα)/(sinα-cosα)=3变形整理,可得sinα=2cosα,或tanα=2
[sin²（2π-α）+cot（3/2*π-α）]/[tan(π+α)(1+sin²α)]
=[sin²α+tanα]/[tanαcos²α]
=[sin²α+2]/[2cos²α]
=（2sin²α+cos²α）/[2cos²α]
=sin²α/cos²α+1/2
=4+1/2
-12/5吧，由已知联系sina的平方+cosa的平方=1求的sina=五分之二倍根五，cosa=五分之根五，再带入后面式子就可以啦
额
题没发完
1/3或7/9