Let α and β be the roots of the quadratic equation x² sinθ - x(sinθ cosθ + 1) + cosθ = 0 (0

Q: Wait, why did we factorize instead of using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$?

Great question! While the quadratic formula always works, it often leads to messy square roots in trigonometric equations. Factorization is a more elegant and faster approach here. By expanding the middle term, we revealed a structure that allowed us to group terms and factorize directly, saving us from complex algebraic manipulation.

Q: How do we know for sure that $\alpha = \cos\theta$ and $\beta = \frac{1}{\sin\theta}$?

We use the given condition $0 1$, it is clear that $\cos\theta < \frac{1}{\sin\theta}$. Thus, $\alpha = \cos\theta$ and $\beta = \frac{1}{\sin\theta}$.

Q: Is the infinite sum formula valid for all values of $\theta$?

The formula $S_{\infty} = \frac{a}{1-r}$ is valid only when $|r| < 1$. In our case, the common ratios are $\cos\theta$ and $-\sin\theta$. Since $0 < \theta < 45^\circ$, we know $0 < \sin\theta < \frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}} < \cos\theta < 1$. Both $|\cos\theta|$ and $|-\sin\theta|$ are strictly less than $1$, so the infinite sum converges perfectly!

Question

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

Options:

(A) $\frac{1}{1-\cos	heta} - \frac{1}{1+\sin	heta}$

(B) $\frac{1}{1+\cos	heta} + \frac{1}{1-\sin	heta}$

(C) $\frac{1}{1+\cos	heta} - \frac{1}{1-\sin	heta}$

(D) $\frac{1}{1-\cos	heta} + \frac{1}{1+\sin	heta}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup The given quadratic equation is: $$x^2 \sin heta - x(\sin heta \cos heta + 1) + \cos heta = 0$$ This equation appears complex, but it can be simplified through strategic factorization. ### The Art of Factorization First, expand the middle term to reveal the structure of the expression: $$x^2 \sin heta - x \sin heta \cos heta - x + \cos heta = 0$$ Next, group the terms to factorize the quadratic: $$x \sin heta (x - \cos heta) - 1(x - \cos heta) = 0$$ This leads to the factored form: $$(x \sin heta - 1)(x - \cos heta) = 0$$ Thus, the roots of the equation are $x = \cos heta$ and $x = \frac{1}{\sin heta}$. ### The Range Trap We are given the constraint $0 < heta < 45^\circ$. In this interval, $\sin heta$ lies between $0$ and $\frac{1}{\sqrt{2}}$, which implies: $$\frac{1}{\sin heta} > \sqrt{2} > 1$$ Conversely, $\cos heta$ lies between $\frac{1}{\sqrt{2}}$ and $1$. Since $\cos heta < 1$ and $\frac{1}{\sin heta} > 1$, we conclude that $\cos heta < \frac{1}{\sin heta}$. Given the condition $\alpha < \beta$, we must assign the roots as follows: $$\alpha = \cos heta, \quad \beta = \frac{1}{\sin heta}$$ ### The Infinite Dance We are tasked with evaluating the infinite sum: $$S = \sum_{n=0}^{\infty} \left( \alpha^n + \frac{(-1)^n}{\beta^n} ight)$$ Substituting our values for $\alpha$ and $\beta$, the expression becomes: $$S = \sum_{n=0}^{\infty} (\cos heta)^n + \sum_{n=0}^{\infty} (-\sin heta)^n$$ Both components are infinite geometric series with the first term $a = 1$. The common ratios are $r_1 = \cos heta$ and $r_2 = -\sin heta$. Since $|r_1| < 1$ and $|r_2| < 1$, we apply the sum formula $S_{\infty} = \frac{a}{1-r}$: $$S = \frac{1}{1 - \cos heta} + \frac{1}{1 - (-\sin heta)}$$ ### Final Result Simplifying the expression, we obtain the final sum: **$$S = \frac{1}{1 - \cos heta} + \frac{1}{1 + \sin heta}$$**

Animated Solution for Mathematics - Sequence and Series: Let $α$ and $β$ be the roots of the quadratic equation $x^{2} sin θ - x (sin θ cos θ + 1) + cos θ = 0$ ( $0 < θ < 4 5^{\circ}$ ), and $α < β$ . Then $\sum_{n = 0}^{\infty} (α^{n} + \frac{( - 1 ) ^{n}}{β ^{n}})$ is equal to:

Select Answer:

Visualized Solution

The Quadratic Equation

Expanding the Middle Term

Grouping Terms

Factorizing the Equation

Finding the Roots

Analyzing the Range

Comparing the Roots

Assigning $α$ and $β$

Setting up the Infinite Sum

Infinite GP Formula

Evaluating the First Series

Evaluating the Second Series

Final Result

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Art of Factorization

The Range Trap

The Infinite Dance

Final Result

Similar Questions

If $α$ and $β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$ , then $lim_{n \to \infty} \sum_{r = 1}^{n} α^{r} + lim_{n \to \infty} \sum_{r = 1}^{n} β^{r}$ is equal to :

If $0 < θ, ϕ < \frac{π}{2}$ , $x = \sum_{n = 0}^{\infty} cos^{2 n} θ$ , $y = \sum_{n = 0}^{\infty} sin^{2 n} ϕ$ and $z = \sum_{n = 0}^{\infty} cos^{2 n} θ \cdot sin^{2 n} ϕ$ then

Let $729, 81, 9, 1, \dots$ be a sequence and $P_{n}$ denote the product of the first $n$ terms of this sequence. If $2 \sum_{n = 1}^{40} (P_{n})^{1/ n} = \frac{3 ^{α} - 1}{3 ^{β}}$ and $g cd (α, β) = 1$ , then $α + β$ is equal to

Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .

If $x = \sum_{n = 0}^{\infty} (- 1)^{n} tan^{2 n} θ$ and $y = \sum_{n = 0}^{\infty} cos^{2 n} θ$ , for $0 < θ < \frac{π}{4}$ , then:

If $2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + .... + 2 \cdot 3^{9} + 3^{10} = S - 2^{11}$ then $S$ is equal to :

Let the first term $a$ and the common ratio $r$ of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

The sum of first four terms of a geometric progression (G.P.) is $\frac{65}{12}$ and the sum of their respective reciprocals is $\frac{65}{18}$ . If the product of first three terms of the G.P. is 1, and the third term is $α$ , then $2 α$ is

Select Answer:

Visualized Solution

The Quadratic Equation

Expanding the Middle Term

Grouping Terms

Factorizing the Equation

Finding the Roots

Analyzing the Range

Comparing the Roots

Assigning .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Setting up the Infinite Sum

Infinite GP Formula

Evaluating the First Series

Evaluating the Second Series

Final Result

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Art of Factorization

The Range Trap

The Infinite Dance

Final Result

Similar Questions

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Assigning $α$ and $β$

If $α$ and $β$ are the roots of the equation $375 x^{2} - 25 x - 2 = 0$ , then $lim_{n \to \infty} \sum_{r = 1}^{n} α^{r} + lim_{n \to \infty} \sum_{r = 1}^{n} β^{r}$ is equal to :

If $0 < θ, ϕ < \frac{π}{2}$ , $x = \sum_{n = 0}^{\infty} cos^{2 n} θ$ , $y = \sum_{n = 0}^{\infty} sin^{2 n} ϕ$ and $z = \sum_{n = 0}^{\infty} cos^{2 n} θ \cdot sin^{2 n} ϕ$ then

Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .

If $x = \sum_{n = 0}^{\infty} (- 1)^{n} tan^{2 n} θ$ and $y = \sum_{n = 0}^{\infty} cos^{2 n} θ$ , for $0 < θ < \frac{π}{4}$ , then:

If $2^{10} + 2^{9} \cdot 3^{1} + 2^{8} \cdot 3^{2} + .... + 2 \cdot 3^{9} + 3^{10} = S - 2^{11}$ then $S$ is equal to :