tan(a+π/4)=3+2倍根号2,则1-cos2a/sin2a=

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tan(a+π/4)=3+2倍根号2,则1-cos2a/sin2a=

∵tan(a+π/4)=3+2√2 ==>(1+tana)/(1-tana)=3+2√2 (应用正切和角公式展开)
==>2/(1-tana)=4+2√2 (两端同时加1)
==>1-tana=1/(2+√2)
==>tana=1-1/(2+√2)
==>tana=√2/2
∴1-cos2a/sin2a=2cos²a/(2sina*cosa) (应用正余弦半角公式)
=cosa/sina
=tana
=√2/2.

tan(a+π/4)=(tana+1)/(1-tana)=3+2√2
-1 +2/(1-tana)=3+2√2
1/(1-tana)=2+√2
1-tana=1-√2/2
tana=√2/2
(1-cos2a)/sin2a=2(sina)^2/(2sinacosa)=tana=√2/2