Let alpha and beta be the roots of the quadra | vedclass

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(C) The given quadratic equation is $x^2 \sin 	heta - x(\sin 	heta \cos 	heta + 1) + \cos 	heta = 0$.Using the quadratic formula $x = \frac{-b \pm

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$\frac{1}{1 - \cos 	heta} + \frac{1}{1 + \sin 	heta}$

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(C) The given quadratic equation is $x^2 \sin 	heta - x(\sin 	heta \cos 	heta + 1) + \cos 	heta = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we have:$x = \frac{(\sin 	heta \cos 	heta + 1) \pm \sqrt{(\sin 	heta \cos 	heta + 1)^2 - 4 \sin 	heta \cos 	heta}}{2 \sin 	heta}$Since $(\sin 	heta \cos 	heta + 1)^2 - 4 \sin 	heta \cos 	heta = (\sin 	heta \cos 	heta - 1)^2$,we get:$x = \frac{(\sin 	heta \cos 	heta + 1) \pm (\sin 	heta \cos 	heta - 1)}{2 \sin 	heta}$Taking the positive sign: $x = \frac{2 \sin 	heta \cos 	heta}{2 \sin 	heta} = \cos 	heta$.Taking the negative sign: $x = \frac{2}{2 \sin 	heta} = \csc 	heta$.Given $0  \sin 	heta$,so $\cos 	heta The sum is $\sum_{n=0}^\infty \alpha^n + \sum_{n=0}^\infty (-\frac{1}{\beta})^n = \sum_{n=0}^\infty (\cos 	heta)^n + \sum_{n=0}^\infty (-\sin 	heta)^n$.Using the infinite geometric series formula $\sum_{n=0}^\infty r^n = \frac{1}{1-r}$:Sum $= \frac{1}{1 - \cos 	heta} + \frac{1}{1 - (-\sin 	heta)} = \frac{1}{1 - \cos 	heta} + \frac{1}{1 + \sin 	heta}$.

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

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