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