If $\underset{\mathrm{x}\to 0}{lim} {\left(1+\mathrm{xlog}\left(1+{\mathrm{b}}^2\right)\right)}^{1/\mathrm{x}}=2{\mathrm{bsin}}^2\mathrm{\theta }$ , b>0 and  $\mathrm{\theta }\in [-\mathrm{\pi },\mathrm{\pi }]$ then the value of $\theta$ is

$\begin{array}{l}\underset{\mathrm{x}\to 0}{lim} {\left(1+\mathrm{xlog}\left(1+{\mathrm{b}}^2\right)\right)}^{1/\mathrm{x}}\\ =\underset{\mathrm{x}\to 0}{lim} {\left(1+\mathrm{xlog}\left(1+{\mathrm{b}}^2\right)\right)}^{\frac{1}{\mathrm{xlog}\left(1+{\mathrm{b}}^2\right)}\times \log \left(1+{\mathrm{b}}^2\right)}\\ ={\mathrm{e}}^{\log \left(1+{\mathrm{b}}^2\right)}=1+{\mathrm{b}}^2\end{array}$

thus,

$\begin{array}{r}1+{\mathrm{b}}^2=2{\mathrm{bsin}}^2\mathrm{\theta }\\ \Rightarrow {\sin }^2\mathrm{\theta }=\frac{1+{\mathrm{b}}^2}{2\mathrm{b}}\ge 1\end{array}$

${\sin }^2\mathrm{\theta }=1\Rightarrow \sin \mathrm{\theta }=\pm 1\text{ or }\mathrm{\theta }=\pm \mathrm{\pi }/2$