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
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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