Question

If $1+\sin \theta +{\sin }^2\theta +...\mathrm{\infty }=4+2\sqrt{3}$, $0<\theta <\pi$ and $\theta \ne \frac{\pi }{2}$, then $\theta$ is equal to

• A. $\theta =\frac{\pi }{3}$
• B. $\theta =\frac{\pi }{6}$
• C. $\theta =\frac{\pi }{3}or\frac{\pi }{6}$
• D. $\theta =\frac{\pi }{3}or\frac{2\mathrm{\pi }}{3}$

Answer 1

The correct option is D

$\theta =\frac{\pi }{3}or\frac{2\mathrm{\pi }}{3}$

To find the value of $\theta$:

Given, $1+\sin \theta +{\sin }^2\theta +...\mathrm{\infty }=4+2\sqrt{3}$, $0<\theta <\pi$ and $\theta \ne \frac{\pi }{2}$,

$1+\sin \theta +{\sin }^2\theta +...\mathrm{\infty }$ is geometric progression with first term, $a=1$ and common ratio, $r=\sin \theta$ .

As we know,

Sum of geometric progression upto $\infty$, ${\mathit{S}}_{\mathbf{\infty }}\mathbf{=}\frac{\mathbf{a}}{\mathbf{1}\mathbf{-}\mathbf{r}}$,

where $a$ is the first term and $r$ is the common ratio of geometric progression.

$\therefore \begin{array}{rl}1+\sin \theta +{\sin }^2\theta +...\mathrm{\infty }& =\frac{1}{1-\sin \theta }\end{array}$

$\Rightarrow$ $\begin{array}{rl}4+2\sqrt{3}& =\frac{1}{1-\sin \theta }\end{array}$

$\Rightarrow$ $\begin{array}{rl}1-\sin \theta & =\frac{1}{4+2\sqrt{3}}\end{array}$

$\Rightarrow$ $\begin{array}{rl}1-\sin \theta & =\frac{4-2\sqrt{3}}{(4+2\sqrt{3})(4-2\sqrt{3})}\end{array}$

$\begin{array}{rl}& =\frac{4-2\sqrt{3}}{16-12}\\ & =\frac{4-2\sqrt{3}}{4}\\ & =1-\frac{\sqrt{3}}{2}\end{array}$

$\Rightarrow$ $\begin{array}{rl}\sin \theta & =1-1+\frac{\sqrt{3}}{2}\end{array}$

$\Rightarrow$ $\begin{array}{rl}\sin \theta & =\frac{\sqrt{3}}{2}\end{array}$

$\begin{array}{rl}\mathbf{\therefore }\mathbf{\theta }& \mathbf{}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{,}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{3}}\end{array}$

Hence, option (D) is the correct option.