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(D) The given series is an infinite geometric progression with the first term $a = 1$ and common ratio $r = \cos \alpha$.Since the sum of an infinite

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$3\pi / 4$

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(D) The given series is an infinite geometric progression with the first term $a = 1$ and common ratio $r = \cos \alpha$.Since the sum of an infinite geometric series is given by $S = \frac{a}{1 - r}$,we have:$\frac{1}{1 - \cos \alpha} = 2 - \sqrt{2}$$1 - \cos \alpha = \frac{1}{2 - \sqrt{2}}$Rationalizing the denominator:$1 - \cos \alpha = \frac{2 + \sqrt{2}}{(2 - \sqrt{2})(2 + \sqrt{2})} = \frac{2 + \sqrt{2}}{4 - 2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{1}{\sqrt{2}}$$-\cos \alpha = \frac{1}{\sqrt{2}}$$\cos \alpha = -\frac{1}{\sqrt{2}}$Since $0 < \alpha < \pi$,$\alpha = \frac{3\pi}{4}$.

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

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