Question

If $1+\sin \theta+\sin ^2 \theta+\cdots \cdot$ upto $\infty=2 \sqrt{3}+4$, then $\theta=$.............

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{3}$
• D. $\frac{3\pi}{4}$

Answer 1

Answer: (C)

We have, $1+\sin\, \theta+\sin ^{2} \theta+\ldots \infty=2 \sqrt{3}+4$
$\Rightarrow \frac{1}{1-\sin \, \theta}=2 \sqrt{3}+4$
$\left[\because 1+r+r^{2}+\ldots \infty=\frac{1}{1-r}\right]$
$\Rightarrow 1-\sin \, \theta=\frac{1}{2 \sqrt{3}+4}$
$\Rightarrow \sin \,\theta=1-\frac{1}{2 \sqrt{3}+4}$
$\Rightarrow \sin \, \theta=\frac{2 \sqrt{3}+4-1}{2 \sqrt{3}+4}$
$\Rightarrow \sin \, \theta=\frac{2 \sqrt{3}+3}{2 \sqrt{3}+4}=\frac{\sqrt{3}(2+\sqrt{3})}{2(\sqrt{3}+2)}$
$\Rightarrow \sin \,\theta=\frac{\sqrt{3}}{2}$
$\therefore \theta=60^{\circ}=\frac{\pi}{3}$