Question

.《三角恒等变换》1已知a,b在（0,45）范围内,tan(a/2)/1-tan^2(a/2)=1/4,且3sinb=s

Answer 1

2.不知道错没,

sin2a=3/5 =>cos2a=4/5

tan2a=3/4

tan2a=2tan a /1-tan^2 a=3/4

3tan^2 a+8tan a-3=0

(3tan a-1)(tan a+3)=0

tan a=1/3 或tan a=-3（舍）

tan(a-b)=tana -tanb/1+tanatanb

tanb=tanb=-1/7

tan(a+b)=tana+tanb/1-tanatanb=2/11