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If \( 1+\sin \theta+\sin ^{2} \theta+\ldots . \) upto \( \infty=2 \sqrt{3}+4 \), then \( \theta= \)
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Verified Answer
The correct answer is:
\( \frac{\pi}{3} \)
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 $r=\sin \theta$
$$
\begin{array}{l}
\text { So, } \frac{1}{1-\sin \theta}=2 \sqrt{3}+4 \\
\Rightarrow 1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \\
\Rightarrow 1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \times \frac{2 \sqrt{3}-4}{2 \sqrt{3}-4} \\
\Rightarrow 1-\sin \theta=\frac{2 \sqrt{3}-4}{12-16}=\frac{2 \sqrt{3}-4}{-4} \\
\Rightarrow 1-\sin \theta=\frac{-\sqrt{3}}{2}+1 \\
\Rightarrow 1-\sin \theta=\frac{-\sqrt{3}}{2}+1 \\
\Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{array}
$$
$1+\sin \theta+\sin ^{2} \theta+\ldots+\infty=2 \sqrt{3}+4$
We know that $S_{\infty}=\frac{a}{1-r}$
Here $a=1$ and $r=\sin \theta$
$$
\begin{array}{l}
\text { So, } \frac{1}{1-\sin \theta}=2 \sqrt{3}+4 \\
\Rightarrow 1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \\
\Rightarrow 1-\sin \theta=\frac{1}{2 \sqrt{3}+4} \times \frac{2 \sqrt{3}-4}{2 \sqrt{3}-4} \\
\Rightarrow 1-\sin \theta=\frac{2 \sqrt{3}-4}{12-16}=\frac{2 \sqrt{3}-4}{-4} \\
\Rightarrow 1-\sin \theta=\frac{-\sqrt{3}}{2}+1 \\
\Rightarrow 1-\sin \theta=\frac{-\sqrt{3}}{2}+1 \\
\Rightarrow \sin \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{array}
$$
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