SATHEE: Trigonometrical Equations 1 Question 5

Trigonometrical Equations 1 Question 5

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

(a) $\frac{1}{1 - \cos θ} - \frac{1}{1 + \sin θ}$

(b) $\frac{1}{1 - \cos θ} + \frac{1}{1 + \sin θ}$

(d) $\frac{1}{1 + \cos θ} + \frac{1}{1 - \sin θ}$

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Answer:

Correct Answer: 5. (b)

Solution:

Given

$x^{2} \sin θ - x \sin θ \cos θ - x + \cos θ = 0$ ,

where $0 < θ < 45^{\circ}$

$\Rightarrow x \sin θ (x - \cos θ) - 1 (x - \cos θ) = 0$

$\Rightarrow$

$(x - \cos θ) (x \sin θ - 1) = 0$

$\begin{aligned} \Rightarrow x = \cos θ, x = cosec θ \\ \Rightarrow α = \cos θ and β = cosec θ \\ (∵ For 0 < θ < 45^{\circ}, \frac{1}{\sqrt{2}} < \cos θ < 1 and \sqrt{2} < cosec θ < \infty \\ \Rightarrow \cos θ < cosec θ) \end{aligned}$

Now, consider, $\sum_{n = 0}^{\infty} α^{n} + \frac{(- 1)^{n}}{β^{n}} = \sum_{n = 0}^{\infty} α^{n} + \sum_{n = 0}^{\infty} \frac{(- 1)^{n}}{β^{n}}$

$= (1 + α + α^{2} + α^{3} + \dots \infty)$

$+ 1 - \frac{1}{β} + \frac{1}{β^{2}} - \frac{1}{β^{3}} + \dots \infty$

$\begin{aligned} = \frac{1}{1 - α} + \frac{1}{1 - \frac{1}{β}} = \frac{1}{1 - α} + \frac{1}{1 + \frac{1}{β}} \\ = \frac{1}{1 - \cos θ} + \frac{1}{1 + \sin θ} ∵ \frac{1}{β} = \sin θ \end{aligned}$

Trigonometrical Equations 1 Question 5

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

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Trigonometrical Equations 1 Question 5

5. Let α and β be the roots of the quadratic equation x2sin⁡θ−x(sin⁡θcos⁡θ+1)+cos⁡θ=0(0<θ<45∘) and α<β. Then, ∑n=0∞αn+(−1)nβn is equal to

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