Showing $8(\arctan(\sqrt2-1)) = \pi$ My friend is saying  $$8(\arctan(\sqrt2-1)) = \pi$$
I know $4\arctan(1) = \pi$
I thought of $4\arctan(\sqrt2-1) + 4\arctan(\sqrt2-1)$
But I got stuck somewhere
How to solve?

After taking the $ \tan $ on both sides you get:
$$ \tan\frac{\pi}{8} = \sqrt{2}-1 $$
Note these two double angle formulas $ \sin2x = 2\sin x \cos x $ and $ \cos 2x = 2\cos^2x-1 $. Now we have:
$$ \tan\frac{\pi}{8} = \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}} =  \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}}  \cdot \frac{\cos\frac{\pi}{8}}{\cos\frac{\pi}{8}} = \frac{\sin\frac{\pi}{4}}{1+\cos\frac{\pi}{4}} = \frac{\sqrt{2}/2}{1+\sqrt{2}/2} = \sqrt{2} - 1$$
