Question

Let $\alpha $ and $\beta $ be the roots of the equation ${{x}^{2}}\operatorname{Sin}\theta -x\left( \operatorname{Sin}\theta \operatorname{Cos}\theta +1 \right)+\operatorname{Cos}\theta =0$
$\left( 0<\theta <{{45}^{\circ }} \right)$ , and $\alpha <\beta $ . Then $\sum\limits_{n=0}^{\infty }{\left( {{\alpha }^{n}}+\dfrac{{{\left( -1 \right)}^{n}}}{{{\beta }^{n}}} \right)}$ is equal to:
(a) $\dfrac{1}{1-\operatorname{Cos}\theta }+\dfrac{1}{1+\operatorname{Sin}\theta }$
(b) $\dfrac{1}{1+\operatorname{Cos}\theta }+\dfrac{1}{1-\operatorname{Sin}\theta }$
(c) $\dfrac{1}{1-\operatorname{Cos}\theta }-\dfrac{1}{1+\operatorname{Sin}\theta }$
(d) $\dfrac{1}{1+\operatorname{Cos}\theta }-\dfrac{1}{1-\operatorname{Sin}\theta }$

Answer 1

Hint: Find the roots of the given equation, put the value of $\alpha $ and $\beta $ in the summation .( Choose the value of $\alpha $ and $\beta $ by keeping in mind that $\alpha <\beta $ and $\left( 0<\theta <{{45}^{\circ }} \right)$ ). Now separate the summation of both the terms and observe that they are Geometric Series with some common ratio.
And we know that the infinite sum of Geometric series is equal to $\dfrac{1}{1-CommonRatio}$ .

Complete step-by-step answer:
To find the summation , we need the values of $\alpha $ and $\beta $ , so for that we will first find the roots of the given quadratic equation,
Consider the given quadratic equation,
${{x}^{2}}\operatorname{Sin}\theta -x\left( \operatorname{Sin}\theta \operatorname{Cos}\theta +1 \right)+\operatorname{Cos}\theta =0$
$\begin{align}
  & x=\dfrac{\left( \operatorname{Sin}\theta \operatorname{Cos}\theta +1 \right)\pm \sqrt{{{\left( \operatorname{Sin}\theta \operatorname{Cos}\theta +1 \right)}^{2}}-4\operatorname{Sin}\theta \operatorname{Cos}\theta }}{2\operatorname{Sin}\theta } \\
 & =\dfrac{\left( \operatorname{Sin}\theta \operatorname{Cos}\theta +1 \right)\pm \sqrt{\left( {{\operatorname{Sin}}^{2}}\theta {{\operatorname{Cos}}^{2}}\theta +1 \right)-2\operatorname{Sin}\theta \operatorname{Cos}\theta }}{2\operatorname{Sin}\theta } \\
 & =\dfrac{\left( \operatorname{Sin}\theta \operatorname{Cos}\theta +1 \right)\pm \sqrt{{{\left( \operatorname{Sin}\theta \operatorname{Cos}\theta -1 \right)}^{2}}}}{2\operatorname{Sin}\theta } \\
 & =\dfrac{\left( \operatorname{Sin}\theta \operatorname{Cos}\theta +1 \right)\pm \left( \operatorname{Sin}\theta \operatorname{Cos}\theta -1 \right)}{2\operatorname{Sin}\theta } \\
 & =\operatorname{Cos}\theta ,\operatorname{Cosec}\theta
\end{align}$
Now that we have the roots of the given equation, we have to decide which is $\alpha $ and which is $\beta $ based on the information that
$\left( 0<\theta <{{45}^{\circ }} \right)$ and $\alpha <\beta $
In this range Cos is always between 0 and 1, Cosec is always between $\sqrt{2}$ and $\infty $
Hence , in this range Cosec is always greater than Cos
Hence , $\alpha =\operatorname{Cos}\theta $ and \[\beta =\operatorname{Cosec}\theta \]
Now, the given summation becomes,
$\sum\limits_{n=0}^{\infty }{\left( {{\operatorname{Cos}}^{n}}\theta +\dfrac{{{\left( -1 \right)}^{n}}}{{{\left( \operatorname{Cosec}\theta \right)}^{n}}} \right)}$
Now, we will separate it into 2 sums , and get,
$\sum\limits_{n=0}^{\infty }{\left( {{\operatorname{Cos}}^{n}}\theta \right)+}\sum\limits_{n=0}^{\infty }{\dfrac{{{\left( -1 \right)}^{n}}}{{{\left( \operatorname{Cosec}\theta \right)}^{n}}}}$
Now, Since $\left| \operatorname{Cos}\theta \right|\le 1$
so the first geometric infinite series with common ratio $\operatorname{Cos}\theta $ and since the common ratio is less than 1 , this series is convergent and converges to $\dfrac{1}{1-\operatorname{Cos}\theta }$ .
and the second series is also a geometric series with common ratio $-\dfrac{1}{\operatorname{Cosec}\theta }$
Now, we will check if this ratio is less than 1 or not
$\sqrt{2}<\operatorname{Cosec}\theta <\infty $
Hence , $0<\dfrac{1}{\operatorname{Cosec}\theta }<\dfrac{1}{\sqrt{2}}$
Hence , second series has the common ratio less than 1
And hence it is also convergent and converges to $\begin{align}
  & \dfrac{1}{1-\left( -\dfrac{1}{\operatorname{Cosec}\theta } \right)} \\
 & =\dfrac{1}{1+\operatorname{Sin}\theta } \\
\end{align}$
Hence , the whole series converges to $\dfrac{1}{1-\operatorname{Cos}\theta }+\dfrac{1}{1+\operatorname{Sin}\theta }$ .

So, the correct answer is “Option A”.

Note: The possibility of mistake here is that students can forget the sign of the common ratio of the second series . The series also has ${{\left( -1 \right)}^{n}}$ , so its common ratio has negative signs in it . So, we have to take the common ratio to be the term $-\dfrac{1}{\operatorname{Cosec}\theta }$ and evaluate the sum of the geometric series.
And get the desired sum.