ধরা যাক \( \alpha \) এবং \( \beta \) দ্বিঘাত সমীকরণ \( x^2 \sin \theta - x (\sin \theta \cos \theta + 1) + \cos \theta = 0 \) এর বীজ।

\( (0 < \theta < 45^\circ) \)

এবং \( \alpha < \beta \)। তাহলে \( \sum_{n=0}^{\infty} \left( a^n + \dfrac{(-1)^n}{\beta^n} \right) \) এর মান হবে:

1
\( \dfrac{1}{1 - \cos \theta} + \dfrac{1}{1 + \sin \theta} \)
2
\( \dfrac{1}{1 + \cos \theta} + \dfrac{1}{1 - \sin \theta} \)
3
\( \dfrac{1}{1 - \cos \theta} - \dfrac{1}{1 + \sin \theta} \)
4
\( \dfrac{1}{1 + \cos \theta} - \dfrac{1}{1 - \sin \theta} \)

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