Question

If \[1 + \sin \theta + {\sin ^2}\theta + ..... = 4 + 2\sqrt 3 \], \[0 < \theta < \pi,\theta \ne \dfrac{\pi }{2}\] then
A. \[\theta = \dfrac{\pi }{3}\]
B. \[\theta = \dfrac{\pi }{6}\]
C. \[\theta = \dfrac{\pi }{3}or\dfrac{\pi }{6}\]
D. \[\theta = \dfrac{\pi }{3}or\dfrac{{2\pi }}{3}\]

Answer 1

Hint: When we observe the given series we come to know it is related to one trigonometric function that is sine. But it is along with a combination of geometric progression with the common ratio as \[\sin \theta \]. We will use the formula of G.P. to find the sum of the series on LHS and then the actual sum is already given. We will equate them to solving it and we will get the values of \[\theta \]. Then we will decide the answer.

Complete step by step answer:
Given the series is \[1 + \sin \theta + {\sin ^2}\theta + ..... = 4 + 2\sqrt 3 \]
We can observe that, it is a G.P. with \[\sin \theta \] as the common ratio.
We know that the sum of terms in a G.P. is given by,
\[S = \dfrac{a}{{1 - r}}\]
Here a is the first term equals to 1 and r is the common ratio equals to \[\sin \theta \]. Putting these values,
\[S = \dfrac{1}{{1 - \sin \theta }}\]
But the sum is already given to us, then we will equate them as;
\[\dfrac{1}{{1 - \sin \theta }} = 4 + 2\sqrt 3 \]
On transposing the terms,
\[1 - \sin \theta = \dfrac{1}{{4 + 2\sqrt 3 }}\]
Multiplying and dividing the RHS by \[4 - 2\sqrt 3 \] ,
\[1 - \sin \theta = \dfrac{1}{{4 + 2\sqrt 3 }} \times \dfrac{{4 - 2\sqrt 3 }}{{4 - 2\sqrt 3 }}\]
The denominator now becomes,
\[1 - \sin \theta = \dfrac{{4 - 2\sqrt 3 }}{{{4^2} - {{\left( {2\sqrt 3 } \right)}^2}}}\]
Taking the squares,
\[1 - \sin \theta = \dfrac{{4 - 2\sqrt 3 }}{{16 - 12}}\]
On subtracting we get,
\[1 - \sin \theta = \dfrac{{4 - 2\sqrt 3 }}{4}\]
Separating the terms,
\[1 - \sin \theta = 1 - \dfrac{{\sqrt 3 }}{2}\]
Cancelling 1 from both sides,
\[\sin \theta = \dfrac{{\sqrt 3 }}{2}\]
To find the value of \[\theta \]
\[\theta = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)\]
We know that, \[\sin \dfrac{{n\pi }}{3} = \dfrac{{\sqrt 3 }}{2}\]
\[\therefore \theta = \dfrac{\pi }{3}\,or\,\dfrac{{2\pi }}{3}\] this is because \[0 < \theta < \pi ,\theta \ne \dfrac{\pi }{2}\]

Thus option D is the correct answer.

Note:Here we can get confused about the options. But we should check the limit or restrictions on the \[\theta \] . It is greater than 0 but less than \[\pi \]. Talking about \[\dfrac{{2\pi }}{3}\] it lies below or under the limit so assigned. So choose options wisely.
Second thing that is to be noted is the G.P. and the common ratio $r$. A common ratio is obtained when we take the ratio of two consecutive terms of a G.P.