If $1+\cos\alpha+\cos^2\alpha+\cos^3\alpha......\infty = 2- \sqrt 2$, then $ \alpha$ =? $(0<\alpha<\pi)$

Question

The series is clearly a GP, so it’s sum is $$2-\sqrt 2=\frac{1}{1-\cos \alpha}$$ $$\cos\alpha=\frac{1-\sqrt 2}{2-\sqrt 2}$$

Simplification is where I am stuck. The answer is $\frac{3\pi}{4}$, so $\cos\alpha=-\frac{1}{\sqrt 2}$

How should I get it till there?

Answer 1

$$\cos\alpha=\frac{(1-\sqrt2)(2+\sqrt2)}{(2-\sqrt2)(2+\sqrt2)} =\frac{-\sqrt2}2=-\frac1{\sqrt2}$$ etc.

Answer 2

Hint

$$ \frac{1-\sqrt 2}{2-\sqrt 2}=\frac{1-\sqrt 2}{2-\sqrt 2}\cdot \frac{2+\sqrt{2}}{2+\sqrt{2}}$$

Answer 3

Set $\sqrt2=a,2=a^2$

$$\dfrac{2-\sqrt2}{1-\sqrt2}=\dfrac{a^2-a}{1-a}=\dfrac{a(a-1)}{-(a-1)}=?$$ as $a-1\ne0$