Question

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^2\sin \theta -x(\sin \theta \cos \theta +1)+\cos \theta =0(0<\theta <{45}^{\circ })$, and $\alpha <\beta$. Then
$\sum _{\infty }^{n=0}({\alpha }^n+\frac{(-1{)}^n}{{\beta }^n})$ is equal to :

• A. $\frac{1}{1+\cos \theta }-\frac{1}{1-\sin \theta }$
• B. $\frac{1}{1-\cos \theta }-\frac{1}{1+\sin \theta }$
• C. $\frac{1}{1+\cos \theta }+\frac{1}{1-\sin \theta }$
• D. $\frac{1}{1-\cos \theta }+\frac{1}{1+\sin \theta }$

Answer 1

The correct option is D $\frac{1}{1-\cos \theta }+\frac{1}{1+\sin \theta }$

$x^2\sin \theta -x(\sin \theta \cos \theta +1)+\cos \theta =0$

$\therefore x=\frac{(\sin \theta \cos \theta +1)\pm \sqrt{(\sin \theta \cos \theta +1{)}^2-4\sin \theta \cos \theta }}{2\sin \theta }$

$\Rightarrow x=\frac{(\sin \theta \cos \theta +1)\pm (\sin \theta \cos \theta -1)}{2\sin \theta }$

$\therefore \alpha =\frac{\sin \theta \cos \theta +1+\sin \theta \cos \theta -1}{2\sin \theta }=\cos \theta$

and $\beta =\frac{\sin \theta \cos \theta +1-\sin \theta \cos \theta +1}{2\sin \theta }\Rightarrow \beta =\frac{1}{\sin \theta }(\beta >\alpha )$

Now, $\sum _{n=0}^{\infty }({\alpha }^n+\frac{(-1{)}^n}{{\beta }^n})$
$=\sum _{n=0}^{\infty }{\alpha }^n+\sum _{n=0}^{\infty }\frac{(-1{)}^n}{{\beta }^n}$
$=\sum _{n=0}^{\infty }(\cos \theta {)}^n+\sum _{n=0}^{\infty }(-\sin \theta {)}^n=\frac{1}{1-\cos \theta }+\frac{1}{1+\sin \theta }$