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JEE Main 2019 · 11 Jan · Shift II

JEE Main 2019 Mathematics question on Complex Numbers and Quadratic Equations

Let α\alpha and β\beta be the roots of the quadratic equation x2sin⁡θ−x(sin⁡θcos⁡θ+1)+cos⁡θ=0x^2 \sin\theta - x(\sin\theta\cos\theta + 1) + \cos\theta = 0 (0<θ<45°)(0 < \theta < 45°) and α<β\alpha < \beta. Then ∑n=0∞(αn+(−1)nβn)\sum_{n=0}^{\infty} \left(\alpha^n + \frac{(-1)^n}{\beta^n}\right) is equal to
  1. A.
    11−cos⁡θ+11+sin⁡θ\frac{1}{1-\cos\theta} + \frac{1}{1+\sin\theta}
  2. B.
    11−cos⁡θ−11+sin⁡θ\frac{1}{1-\cos\theta} - \frac{1}{1+\sin\theta}
  3. C.
    11+cos⁡θ+11−sin⁡θ\frac{1}{1+\cos\theta} + \frac{1}{1-\sin\theta}
  4. D.
    11+cos⁡θ−11−sin⁡θ\frac{1}{1+\cos\theta} - \frac{1}{1-\sin\theta}
Show answer

Correct option: B

[JEE Main 2019 · 11 Jan · Shift II]

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