Question

Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^{2}sin\theta -x(sin\theta cos\theta +1)+cos\theta =0$
$(0<\theta <{45}^o)$, and $\alpha <\beta$. Then $\stackrel{\infty }{\underset{n=0}{\Sigma }}$ $(a^n+\frac{(-1{)}^n}{{\beta }^n})$ 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

Answer: (A)

$\frac{1}{1-cos\theta }+\frac{1}{1+sin\theta }$

$D=(1+sin\theta cos\theta {)}^2-4sin\theta cos\theta {)}^2$
= roots are$\beta =cos\theta$ and $\alpha =cos\theta$
$\Rightarrow \stackrel{\infty }{\underset{n=0}{\Sigma }}$ $(a^n+\frac{(-1{)}^n}{{\beta }^n})$ $=\stackrel{\infty }{\underset{n=0}{\Sigma }}(cos\theta {)}^n+\stackrel{\infty }{\underset{n=0}{n}}(-sin\theta {)}^n$
$=\frac{1}{1-cos\theta }+\frac{1}{1+sin\theta }$