(sinα-cosα)／(sinα+2cosα+cos²α)
=(tanα-1)／(tanα+2+cosα)
= 1/(4+cosα)
= 1/(4± 1/√5)
=√5/(4√5 ± 1 )cos²α不是分母sinα-cosα)／(sinα+2cosα)+cos²α=(tanα-1)／(tanα+2) +cos²α=1/4 + 1/5= 9/20答案是5分之16 我做不出来问题是否这样[ (sinα-cosα) / (sinα+2cosα)]+cos²α[ (sinα+cosα) / (sinα-cosα)]+cos²α[ (sinα+cosα) / (sinα-cosα)]+cos²α=[ (tanα+1) / (tanα-1)]+cos²α= 3 + 1/5= 16/5[ (tanα+1) / (tanα-1)]+cos²α,然后呢，详细点tanα =2cosα = ±1/√5cos²α = 1/5=[ (tanα+1) / (tanα-1)]+cos²α=(2+1)/(2-1) + 1/5= 3+ 1/5= 16/5