Trigonometrical Equations 1 Question 5
5. Let $\alpha$ and $\beta$ be the roots of the quadratic equation $x^{2} \sin \theta-x(\sin \theta \cos \theta+1)+\cos \theta=0\left(0<\theta<45^{\circ}\right)$ and $\alpha<\beta$. Then, $\sum _{n=0}^{\infty} \alpha^{n}+\frac{(-1)^{n}}{\beta^{n}}$ 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) $\frac{1}{1+\cos \theta}+\frac{1}{1-\sin \theta}$
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Answer:
Correct Answer: 5. (b)
Solution:
- Given
$x^{2} \sin \theta-x \sin \theta \cos \theta-x+\cos \theta=0$,
where $0<\theta<45^{\circ}$
$\Rightarrow x \sin \theta(x-\cos \theta)-1(x-\cos \theta)=0$
$\Rightarrow$
$$ (x-\cos \theta)(x \sin \theta-1)=0 $$
$$ \begin{aligned} & \Rightarrow \quad x=\cos \theta, x=\operatorname{cosec} \theta \\ & \Rightarrow \quad \alpha=\cos \theta \text { and } \beta=\operatorname{cosec} \theta \\ & \left(\because \text { For } 0<\theta<45^{\circ}, \frac{1}{\sqrt{2}}<\cos \theta<1 \text { and } \sqrt{2}<\operatorname{cosec} \theta<\infty\right. \\ & \Rightarrow \quad \cos \theta<\operatorname{cosec} \theta) \end{aligned} $$
Now, consider, $\sum _{n=0}^{\infty} \alpha^{n}+\frac{(-1)^{n}}{\beta^{n}}=\sum _{n=0}^{\infty} \alpha^{n}+\sum _{n=0}^{\infty} \frac{(-1)^{n}}{\beta^{n}}$
$$ =\left(1+\alpha+\alpha^{2}+\alpha^{3}+\ldots \infty\right) $$
$$ +1-\frac{1}{\beta}+\frac{1}{\beta^{2}}-\frac{1}{\beta^{3}}+\ldots \infty $$
$$ \begin{aligned} & =\frac{1}{1-\alpha}+\frac{1}{1–\frac{1}{\beta}}=\frac{1}{1-\alpha}+\frac{1}{1+\frac{1}{\beta}} \\ & =\frac{1}{1-\cos \theta}+\frac{1}{1+\sin \theta} \quad \because \frac{1}{\beta}=\sin \theta \end{aligned} $$
