Question

已知tan(兀/4+a)=2，求1/(2sinacosa+cos平方a )的值

Answer 1

tan(π/4+a)=[tanπ/4+tana)/[1-tanπ/4*tana]=2
(1+tana)/(1-tana)=2
tana=1/3
1/(2sinacosa+cos²a )
=(sin²a+cos²a)/(2sinacosa+cos²a)
=(tan²a+1)/(2tana+1)
=[(1/3)²+1]/(2*1/3+1)
=10/9*3/5
=2/3

Answer 2

tan(π/4+a)=(1+tana)/(1-tana)=2
tana=1/3
1/(2sinacosa+cos²a)
=(1/cos²a)/(2tana+1)
=(1+tan²a)/(2tana+1)
=2/3
望采纳，谢谢！