题目

已知tan(a+b)=1/5,tan(b+π/4)=1/4,求tana的值
1求tana的值
2求sina的平方+sinacosa+cosa的平方

提问时间：2020-07-26

答案

解(1)由tan(a-π/4)=tan[(a+b)-(b+π/4)]
即(tana-1)/(1+tana)=[tan(a+b)-tan(b+π/4)]/[1+tan(a+b)tan(b+π/4)]
即(tana-1)/(1+tana)=(1/5-1/4)/[1+1/4×1/5]
即(tana-1)/(1+tana)=-1/21
即-1-tana=21tana-21
即22tana=20
即tana=10/11
(2)sin^2a+sinacosa+cos^2a
=(sin^2a+sinacosa+cos^2a)/(sin^2a+cos^2a)
=(sin^2a+sinacosa+cos^2a)×1/cos^2a/(sin^2a+cos^2a)×1/cos^2a
=(tan^2a+tana+1)/(tan^2a+1)
=1+tana/(tan^2a+1)
=1+10/11/((10/11)^2+1)
=1+10/11/(100+121)/121
=1+10/(221/11)
=1+110/221
=331/221.