Mathematics - Quadratic Equation Question with Solution | TestHub

$x^2\sin \theta -x(\sin \theta \cdot \cos \theta +1)+\cos \theta =0$. $x^2\sin \theta -x\sin \theta \cdot \cos \theta -x+\cos \theta =0$ $x\sin \theta (x-\cos \theta )-1(x-\cos \theta )=0$ $(x-\cos \theta )(x\sin \theta -1)=0$ $\therefore x=\cos \theta ,\mathrm{cosec}\theta ,\theta \in \left(0,45^{\circ }\right)$ $\therefore \alpha =\cos \theta ,\beta =\mathrm{cosec}\theta$ ${\sum }_{n=0}^{\infty }{\alpha }^n=1+\cos \theta +{\cos }^2\theta +\dots \infty =\frac{1}{1-\cos \theta }$ ${\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{{\beta }^n}=1-\frac{1}{\mathrm{cosec}\theta }+\frac{1}{{\mathrm{cosec}}^2\theta }-\frac{1}{{\mathrm{cosec}}^3\theta }+\dots \infty$ $=1-\sin \theta +{\sin }^2\theta -{\sin }^3\theta +\dots \infty$ $=\frac{1}{1+\sin \theta }$ $\therefore {\sum }_{n=0}^{\infty }\left({\alpha }^n+\frac{(-1{)}^n}{{\beta }^n}\right)={\sum }_{n=0}^{\infty }{\alpha }^n+{\sum }_{n=0}^{\infty }\frac{(-1{)}^n}{{\beta }^n}$ $=\frac{1}{1-\cos \theta }+\frac{1}{1+\sin \theta }.$