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