已知tana⼀2=2求tan(a+π⼀4)

解：
1.
∵tan(a/2)=2
∴tana=[2tan(a/2)]/{1-[tan(a/2)]^2}=(2×2)/(1-2^2)=-4/3
∴tan(a+π/4)=[tana+tan(π/4)]/[1-tana×tan(π/4)]=(-4/3+1)/(1+4/3)=-1/7
2.原式
=(6sina/cosa+cosa/cosa)/(3sina/cosa-2cosa/cosa)
=(6tana+1)/(3tana-2)
=(-6×4/3+1)/(-3×4/3-2)
=7/6

已知tana/2=2求tan(a+π/4)
tana=2tana/2/(1-tan^2a/2)
=4/(1-4)
=-4/3

tan(a+π/4)=(tana+tanπ/4)/(1-tanatanπ/4)
=(-4/3+1)/(1+4/3)
=-1/7
2、sina=2tana/2/(1+tan^2a/2)
=4/(1+4)
=4/5
cosa=(1-tan^2a/2)/(1+tan^2a/2)
=(1-4)/(1+4)
=-3/5
所以有：
(6sina+cosa）/(3sina-2cosa)
=(24/5-3/5)/(12/5+6/5)
=21/18
=7/6