If \( 1+\sin \theta+\sin ^{2} \theta+\ldots . \) upto \( \infty=2 \sqrt{3}+4 \), then \( \theta= \)

Given that,
\(1+\sin \theta+\sin ^{2} \theta+\ldots+\infty=2 \sqrt{3}+4\)
We know that \(S_{\infty}=\frac{a}{1-r}\)
Here \(a=1\) and \(r=\sin \theta\)
\[
\begin{array}{l}
\text { So, } \frac{1}{1-\sin \theta}=2 \sqrt{3}+4 \\
\Rightarrow 1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \\
\Rightarrow 1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \times \frac{2 \sqrt{3}-4}{2 \sqrt{3}-4} \\
\Rightarrow 1-\sin \theta=\frac{2 \sqrt{3}-4}{12-16}=\frac{2 \sqrt{3}-4}{-4} \\
\Rightarrow 1-\sin \theta=\frac{-\sqrt{3}}{2}+1 \\
\Rightarrow 1-\sin \theta=\frac{-\sqrt{3}}{2}+1 \\
\Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{array}
\]