• # question_answer If $1+\cos \alpha +{{\cos }^{2}}\alpha +.......\,\infty =2-\sqrt{2,}$ then $\alpha ,$ $(0<\alpha <\pi )$ is    [Roorkee 2000; AMU 2005] A) $\pi /8$ B) $\pi /6$ C) $\pi /4$ D) $3\pi /4$

$1-\cos \alpha =\frac{1}{2-\sqrt{2}}=1+\frac{1}{\sqrt{2}}$ Þ$\cos \alpha =-\frac{1}{\sqrt{2}}=\cos \frac{3\pi }{4}$ Þ $\alpha =\frac{3\pi }{4}$.