(1-tanα)/(1+tanα)=3+2√2
(1-tanα)/(1+tanα)=2+2√2+1
(1-tanα)/(1+tanα)=(√2+1)^2=(√2+1)/(√2-1)
(1+tanα)(√2+1)=(1-tanα)(√2-1)
(√2+1)+(√2+1)tanα=(√2-1)-(√2-1)tanα
(√2+1+√2-1)tanα=√2-1-√2-1
2√2tanα=-2
tanα=-√2/2
(tanα)^2=(sinα/cosα)^2=(1-(cosα)^2)/(cosα)^2=1/(cosα)^2-1
cosα=±1/√(1+(tanα)^2)
（sinα+cosα)-1/(cotα-sinα*cosα)
=（tanα*cosα+cosα)-1/(cosα/sinα-sinα*cosα)
= ±（1+tan)/√(1+(tanα)^2)-sinα/(cosα-cosα(sinα)^2)
=±（1+tan)/√(1+(tanα)^2)-sinα/(cosα(1-(sinα)^2)
= ±（1+tan)/√(1+(tanα)^2)-tanα*1/(cosα)^2
=±（1+tan)/√(1+(tanα)^2)-tanα(1+(tanα)^2)
=±(1-√2/2)/√(1+(-√2/2)^2)+√2/2(1+(-√2/2)^2)
=±(1-√2/2)/√(3/2)+√2/2(3/2)
=±√(3/2)(1-√2/2)2/3+3√2/4
=±(√6-√3)/3+3√2/4
因?yàn)棣恋闹挡恢嵌嗌?cosα可正可負(fù).所以cosα開(kāi)方修改后加上±.