If 1 + + ^2 + ....... , = 2 - 2, then , (0 < < ) is - Vidyadip

MCQ

If $1 + \cos \alpha + {\cos ^2}\alpha + .......\,\infty = 2 - \sqrt {2,} $ then $\alpha ,$ $(0 < \alpha < \pi )$ is

A
$\pi /8$
B
$\pi /6$
C
$\pi /4$
✓
$3\pi /4$

✓

Answer

Correct option: D.

$3\pi /4$

d
(d) $1 - \cos \alpha = \frac{1}{{2 - \sqrt 2 }} = 1 + \frac{1}{{\sqrt 2 }}$

==> $\cos \alpha = - \frac{1}{{\sqrt 2 }} = \cos \frac{{3\pi }}{4}$

==> $\alpha = \frac{{3\pi }}{4}$.

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