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 $\frac{\pi }{3}$

$1+\sin \theta +{\sin }^2\theta +....upto\infty =\frac{1}{1-\sin \theta }$
LHS is a G.P. with firsr term 1 and common ratio $\sin \theta$

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

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