please ans this question as soon as possible

Hello student,
Given that sin A = 3 sin (A + 2B)
So, sin (A + 2B) / sin A = 1/3
By componendo and dividendo,
[sin (A+2B) + sin A]/ [sin (A+2B) – sin A] = (1+3)/(1-3)
This gives tan (A+B) = – 2 tan B
So, [tan A + tan B]/[1 – tan A tan B] = -2 tan B
So, tan A = 3 tan B/(2 tan²B-1)
< 0 (as A is obtuse)
So, 2 tan ²B -1 < 0 (as tan B > 0
So, 0 < tan B < 1/√2.