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माना द्विघात समीकरण $$ \begin{aligned} x ^{2} \sin \theta- x (\sin \theta \cos \theta+1) &+\cos \theta \\ =& 0\left(0 < \theta < 45^{\circ}\right) \end{aligned} $$ के मूल $\alpha$ तथा $\beta(\alpha<\beta)$ हैं, तो $\sum_{ n =0}^{\infty}\left(\alpha^{ n }+\frac{(-1)^{ n }}{\beta^{ n }}\right)$ बराबर है
$\frac{1}{{1 - \cos \,\theta }} - \frac{1}{{1 + \sin \,\theta \,}}$
$\frac{1}{{1 + \cos \,\theta }} + \frac{1}{{1 - \sin \,\theta \,}}$
$\frac{1}{{1 - \cos \,\theta }} + \frac{1}{{1 + \sin \,\theta \,}}$
$\frac{1}{{1 + \cos \,\theta }} - \frac{1}{{1 - \sin \,\theta \,}}$
Solution
Using quadratic formula,
$x=\frac{(\cos \theta \sin \theta+1) \pm \sqrt{(\cos \theta \sin \theta+1)^{2}-4 \sin \theta \cos \theta}}{2 \sin \theta}$
$=\frac{(\cos \theta \sin \theta+1)^{2} \pm(\cos \theta \sin \theta-1)}{2 \sin \theta}$
$ = \cos \,\theta ,\,\cos ec\,\theta $
$\alpha = \cos \,\theta ,\,\beta = \cos ec\,\theta $
$\therefore \,\sum\limits_{n = 0}^\infty {{\alpha ^n}} + \frac{{{{( – 1)}^n}}}{{{\beta ^n}}}$
$ = \sum\limits_{n = 0}^\infty {{{(\cos ec)}^n}} + \sum\limits_{n = 0}^\infty {{{( – \sin \theta )}^n}} $
$=\frac{1}{1-\cos \theta}+\frac{1}{1+\sin \theta}$
$\therefore $ $(C)$ is the correct
