已知tan(a+派/4)=2,求1+2sinacosa/cos^2a-sin^2a
问题描述：
已知tan(a+派/4)=2,求1+2sinacosa/cos^2a-sin^2a
答
rhy
答
tan(a+π/4)=2，
tana=tan[(a+π/4)-π/4]=[tan(a+π/4)-tan(π/4)]/[1+tan(a+π/4)tan(π/4)]=(2-1)/(1+2×1)=1/3，
(1+2sinacosa)/(cos^2a-sin^2a)=(cos^2a+sin^2a+2sinacosa)/(cos^2a-sin^2a)=(1+tan^2a+2tana)/(1-tan^2a)
=[1+(1/3)^2+2×(1/3)]/[1-(1/3)^2]=(16/9)/(8/9)=2
答
tan(a+π/4)=2=(tana+tanπ/4)/(1-tanatanπ/4)=(tana+1)/(1-tana)
tana=1/3
(1+2sinacosa)/(cos^2a-sin^2a)
=(sin^2a+cos^2a+2sinacosa)/(cos^2a-sin^2a)(上下同除cos^2a)
=((tan^2a+1+2tana)/(1-tan^2a)
=(1/9+1+2/3)/(1-1/9)
=2