Here, $\lim _{x \rightarrow 0}\left\{1+x \log \left(1+b^2\right)\right\}^{1 / x}$
$$
\begin{aligned}
& \text { Given, } \quad\left[1^{\infty} \text { from }\right] \\
& \Rightarrow \quad e^{\lim _{x \rightarrow 0}\left\{x \log \left(1+b^2\right)\right\} \cdot \frac{1}{x}} \\
& \Rightarrow \quad e^{\log \left(1+b^2\right)}=(1+b)^2 \\
& \lim _{x \rightarrow 0}\left\{1+x \log \left(1+b^2\right)\right\}^{1 / x}=2 b \sin ^2 \theta \\
& \Rightarrow \quad\left(1+b^2\right)=2 b \sin ^2 \theta \\
& \therefore \quad \sin ^2 \theta=\frac{1+b^2}{2 b} \quad \ldots \text { (ii) }
\end{aligned}
$$
By $A M \geq G M$,
$$
\frac{b+\frac{1}{b}}{2} \geq\left(b \cdot \frac{1}{b}\right)^{1 / 2} \Rightarrow \frac{b^2+1}{2 b} \geq 1
$$
From Eqs. (ii) and (iii), we get $\sin ^2 \theta=1$
$$
\Rightarrow \quad \theta=\pm \frac{\pi}{2} \text { as } \theta \in(-\pi, \pi]
$$