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