已知tanα=2 求(1+sin2α)/cosα

(1+sin2α)/cos2α=(sinα+cosα)^2/(cos^2α-sin^2α)=cos^2α*(1+tanα)^2/[cos^2α*(1-tan^2α)]=(1+tanα)^2/(1-tan^2α)]=
tanα=2代入
原式=-3

sinα/cosα=tanα=2
sinα=2cosα
代入恒等式sin²α+cos²α=1
cos²α=1/5
cosα=±√5/5

sin2α=2sinαcosα=2*2cosα*cosα=4cos²α=4/5
所以原式=(1+4/5)/(±√5/5)
=9/(±√5)
=±9√5/5

tanα=2
sinα/cosα=√(1-cos^α)/cosα=2
2cosα=√(1-cos^α)
4cos^α=1-cos^α
cosα=±√5/5

(1+sin2α)/cosα
=[(sinα+cosα)/cosα]^2/cosα
=(tanα+1)^2/cosα
=±9√5/5

tana=sina/cosa=2
sina=2cosa
(sina)^2=4(cosa)^2
(sina)^2+(cosa)^2=5(cosa)^2
(cosa)^2=1/5
cosa=(+/-)根号5 /5

(1+sin2α)/cosα
=[1+2sinacosa]/cosa
=[1+2*2cosa*cosa]/cosa
=[1+4*1/5]/cosa
=(9/5)/cosa

(1)cosa=根号5 /5时,上式=(9/5)/(根号5 /5)=(9/5)根号5.

(2)cosa=-根号5 /5时,上式=-(9/5)根号5.