JEE Mains · Maths · STD 11 - 4.2 Quadratic equations and inequations
Let \(\alpha \) and \(\beta \) be the roots of the quadratic equation \({x^2}\,\sin \,\theta - x\,\left( {\sin \,\theta \cos \,\,\theta + 1} \right) + \cos \,\theta = 0\,\left( {0 < \theta < {{45}^o}} \right)\) , and \(\alpha < \beta \). Then \(\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + \frac{{{{\left( { - 1} \right)}^n}}}{{{\beta ^n}}}} \right)} \) is equal to
- A \(\frac{1}{{1 - \cos \,\theta }} - \frac{1}{{1 + \sin \,\theta \,}}\)
- B \(\frac{1}{{1 + \cos \,\theta }} + \frac{1}{{1 - \sin \,\theta \,}}\)
- C \(\frac{1}{{1 - \cos \,\theta }} + \frac{1}{{1 + \sin \,\theta \,}}\)
- D \(\frac{1}{{1 + \cos \,\theta }} - \frac{1}{{1 - \sin \,\theta \,}}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{{1 - \cos \,\theta }} + \frac{1}{{1 + \sin \,\theta \,}}\)
Step-by-step Solution
Detailed explanation
Using quadratic formula, \(x=\frac{(\cos \theta \sin \theta+1) \pm \sqrt{(\cos \theta \sin \theta+1)^{2}-4 \sin \theta \cos \theta}}{2 \sin \theta}\) \(=\frac{(\cos \theta \sin \theta+1)^{2} \pm(\cos \theta \sin \theta-1)}{2 \sin \theta}\)…
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