JEE Main 2019 — Mathematics Algebra

$x^{2} \sin \theta-x(\sin \theta \cdot \cos \theta+1)+\cos \theta=0$x2sinθ−x(sinθ⋅cosθ+1)+cosθ=0. $x^{2} \sin \theta-x \sin \theta \cdot \cos \theta-x+\cos \theta=0$x2sinθ−xsinθ⋅cosθ−x+cosθ=0 $x \sin \theta(x-\cos \theta)-1(x-\cos \theta)=0$xsinθ(x−cosθ)−1(x−cosθ)=0 $(x-\cos \theta)(x \sin \theta-1)=0$(x−cosθ)(xsinθ−1)=0 $\therefore x=\cos \theta, \operatorname{cosec} \theta, \theta \in\left(0,45^{\circ}\right)$∴x=cosθ,cosecθ,θ∈(0,45∘) $\therefore \alpha=\cos \theta, \beta=\operatorname{cosec} \theta$∴α=cosθ,β=cosecθ $\displaystyle \sum_{n=0}^{\infty} \alpha^{n}=1+\cos \theta+\cos ^{2} \theta+\ldots \infty=\frac{1}{1-\cos \theta}$n=0∑∞​αn=1+cosθ+cos2θ+…∞=1−cosθ1​ $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^{n}}{\beta^{n}}=1-\frac{1}{\operatorname{cosec} \theta}+\frac{1}{\operatorname{cosec}^{2} \theta}-\frac{1}{\operatorname{cosec}^{3} \theta}+\ldots \infty$n=0∑∞​βn(−1)n​=1−cosecθ1​+cosec2θ1​−cosec3θ1​+…∞ $=1-\sin \theta+\sin ^{2} \theta-\sin ^{3} \theta+\ldots \infty$=1−sinθ+sin2θ−sin3θ+…∞ $=\frac{1}{1+\sin \theta}$=1+sinθ1​ $\displaystyle \therefore \quad \sum_{n=0}^{\infty}\left(\alpha^{n}+\frac{(-1)^{n}}{\beta^{n}}\right)=\sum_{n=0}^{\infty} \alpha^{n}+\sum_{n=0}^{\infty} \frac{(-1)^{n}}{\beta^{n}}$∴n=0∑∞​(αn+βn(−1)n​)=n=0∑∞​αn+n=0∑∞​βn(−1)n​ $=\frac{1}{1-\cos \theta}+\frac{1}{1+\sin \theta} .$=1−cosθ1​+1+sinθ1​.