1. Evaluate the Limit

The given equation is: $\mathop {\lim }\limits_{x \to 0} {[1 + x\ln (1 + {b^2})]^{1/x}} = 2b{\sin ^2}\theta$

The limit on the left-hand side (LHS) is of the indeterminate form $1^\infty$. We can evaluate it using the standard formula: $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} [f(x)-1]g(x)}$ where $\lim_{x \to a} f(x) = 1$ and $\lim_{x \to a} g(x) = \infty$.

In our case, $f(x) = 1 + x\ln(1+b^2)$ and $g(x) = 1/x$. As $x \to 0$, $f(x) \to 1$ and $g(x) \to \infty$.

Applying the formula: $\text{LHS} = e^{\lim_{x \to 0} [ (1 + x\ln(1+b^2)) - 1 ] \cdot \frac{1}{x} }$ $\text{LHS} = e^{\lim_{x \to 0} [ x\ln(1+b^2) ] \cdot \frac{1}{x} }$ $\text{LHS} = e^{\lim_{x \to 0} \ln(1+b^2) }$ Since $\ln(1+b^2)$ is a constant with respect to $x$, the limit is simply the constant itself. $\text{LHS} = e^{\ln(1+b^2)}$ Using the property $e^{\ln(y)} = y$, we get: $\text{LHS} = 1 + b^2$

2. Set up the Trigonometric Equation

Now, we equate the value of the limit with the right-hand side (RHS) of the given equation: $1 + b^2 = 2b\sin^2\theta$ We can rearrange this to solve for $\sin^2\theta$: $\sin^2\theta = \frac{1+b^2}{2b}$

3. Analyze the Equation

We are given that $b > 0$. We can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality for the positive numbers $b$ and $1/b$. $\frac{b + 1/b}{2} \ge \sqrt{b \cdot \frac{1}{b}}$ $\frac{b + 1/b}{2} \ge 1$ $b + \frac{1}{b} \ge 2$ Now let's look at our expression for $\sin^2\theta$: $\sin^2\theta = \frac{1+b^2}{2b} = \frac{1}{2} \left( \frac{1}{b} + b \right)$ Substituting the result from the AM-GM inequality: $\sin^2\theta \ge \frac{1}{2}(2)$ $\sin^2\theta \ge 1$

4. Solve for $\theta$

We know that the range of the function $\sin^2\theta$ is $[0, 1]$. That is, $\sin^2\theta \le 1$ for all real $\theta$.

We have two conditions:

1. From the given equation and AM-GM inequality: $\sin^2\theta \ge 1$
2. From the properties of the sine function: $\sin^2\theta \le 1$

The only value that satisfies both conditions is: $\sin^2\theta = 1$ This implies that $\sin\theta = 1$ or $\sin\theta = -1$.

The given interval for $\theta$ is $(-\pi, \pi]$.

• If $\sin\theta = 1$, then in the interval $(-\pi, \pi]$, the only solution is $\theta = \frac{\pi}{2}$.
• If $\sin\theta = -1$, then in the interval $(-\pi, \pi]$, the only solution is $\theta = -\frac{\pi}{2}$.

Therefore, the possible values of $\theta$ are $\pm \frac{\pi}{2}$.

5. Conclusion

The value of $\theta$ is $\pm \frac{\pi}{2}$. This corresponds to option D.