Question

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

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

Answer 1

The correct option is C $\frac{\pi }{3}$

$1+\sin \theta +{\sin }^2\theta +............\infty =2\sqrt{3}+4$

LHS is an infinite G.P. with common ratio $\sin \theta$ and first term $1$

$\Rightarrow \frac{1}{1-\sin \theta }=2\sqrt{3}+4=2(\sqrt{3}+2)$

$\Rightarrow 1-\sin \theta =\frac{1}{2(2+\sqrt{3})}$

$\Rightarrow 2(1-\sin \theta )=\frac{1}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}$

$\Rightarrow 2-2\sin \theta =\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

$\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}$

$\Rightarrow \theta =\frac{\pi }{3}$