MathematicsMedium129×since 2002Q5134Let α\alphaα and β\betaβ be the roots of the quadratic equation x² sin θ\thetaθ – x(sin θ\thetaθ cos θ\thetaθ + 1) + cos θ\thetaθ = 0 (0 < θ\thetaθ < 45^o), and α\alphaα < β\betaβ. Then ∑n=0∞(αn+(−1)nβn)\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{{{\left( { - 1} \right)}^n}} \over {{\beta ^n}}}} \right)}n=0∑∞(αn+βn(−1)n) is equal to :A11+cosθ+11−sinθ{1 \over {1 + \cos \theta }} + {1 \over {1 - \sin \theta }}1+cosθ1+1−sinθ1B11−cosθ+11+sinθ{1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}1−cosθ1+1+sinθ1C11−cosθ−11+sinθ{1 \over {1 - \cos \theta }} - {1 \over {1 + \sin \theta }}1−cosθ1−1+sinθ1D11+cosθ−11−sinθ{1 \over {1 + \cos \theta }} - {1 \over {1 - \sin \theta }}1+cosθ1−1−sinθ1Check answerSkip