Q. If $1 + \cos{\alpha + \cos^{2}{\alpha + ...\infty = 2 - \sqrt{2},}\ }$then $\alpha,\ (0 < \alpha < \pi)$ is
A) $\frac{\pi}{8}$
B) $\frac{\pi}{6}$
C) $\frac{\pi}{4}$
D) $\frac{3\pi}{4}$
Explanation
(d)
We have,
$1 + \cos{\alpha + \cos^{2}{\alpha + ...}\ to}\ \infty = 2 - \sqrt{2}$
$\Rightarrow \ \frac{1}{1 - \cos\alpha} = 2 - \sqrt{2}$
$\Rightarrow \ 1 - \cos{\alpha = \frac{1}{2 - \sqrt{2}} = 1 + \frac{1}{\sqrt{2}}}$
$\Rightarrow \ \cos{\alpha = - \frac{1}{\sqrt{2}\ }\ \Rightarrow \ \alpha = \frac{3\pi}{4}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.