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

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

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

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

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

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Sequences 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^{0})

, and

α < β

. Then

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

is equal to:

Answer: (C)

Solving the quadratic by the formula,

x = \frac{( s i n θ c o s θ + 1 ) \pm ( s i n θ c o s θ - 1 )}{2 s i n θ}

, which gives

x = cos θ

x = csc θ

. For

0 < θ < 4 5^{\circ}

cos θ < 1 < csc θ

, so

α = cos θ

and

β = csc θ

. Then

\sum_{n = 0}^{\infty} α^{n} = \frac{1}{1 - c o s θ}

and

\sum_{n = 0}^{\infty} \frac{( - 1 ) ^{n}}{β ^{n}} = \sum (- sin θ)^{n} = \frac{1}{1 + s i n θ}

, whose sum is

\frac{1}{1 - c o s θ} + \frac{1}{1 + s i n θ}

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option