Here, $$\mathop {\lim }\limits_{x \to 0} {\{ 1 + x\log (1 + {b^2})\} ^{1/x}}$$ [1^$$\infty$$ from]

$$ \Rightarrow {e^{\mathop {\lim }\limits_{x \to 0} \{ x\log (1 + {b^2})\} \,.\,{1 \over x}}}$$

$$ \Rightarrow {e^{\log (1 + {b^2})}} = {(1 + b)^2}$$ ..... (i)

Given,

$$\mathop {\lim }\limits_{x \to 0} {\{ 1 + x\log (1 + {b^2})\} ^{1/x}} = 2b{\sin ^2}\theta $$

$$ \Rightarrow (1 + {b^2}) = 2b{\sin ^2}\theta $$

$$\therefore$$ $${\sin ^2}\theta = {{1 + {b^2}} \over {2{b^2}}}$$ ..... (ii)

By $$AM \ge GM$$

$${{b + {1 \over b}} \over 2} \ge {\left( {b\,.\,{1 \over b}} \right)^{1/2}} \Rightarrow {{{b^2} + 1} \over {2b}} \ge 1$$ ...... (iii)

From Eqs. (ii) and (iii), we get $${\sin ^2}\theta = 1$$

$$ \Rightarrow \theta = \pm {\pi \over 2}$$ as $$\theta \in ( - \pi ,\pi ]$$