LimitsHard
Question
If 
[1 + ln(1 + b2)]1/x = 2b sin2 θ, b > 0 and θ ε (- π, π], then the value of θ is
[1 + ln(1 + b2)]1/x = 2b sin2 θ, b > 0 and θ ε (- π, π], then the value of θ is Options
A.
B.
C.
D.
Solution
eln(1 + b2) = 2b sin2θ
⇒ sin2 θ =
⇒ sin2 θ = 1 as ≥ 1
⇒ sin2 θ =
⇒ sin2 θ = 1 as
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