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?
 A: So you have to prove $$\tan {\pi\over 8} = \sqrt{2}-1$$
Notice that $$1= \tan (2\cdot {\pi\over 8}) = {2\tan {\pi\over 8}\over 1-\tan ^2{\pi\over 8}}$$
and solve it on $\tan {\pi\over 8}$.
A: 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:
$$ \tan\frac{\pi}{8} = \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}} =  \frac{\sin\frac{\pi}{8}}{\cos\frac{\pi}{8}}  \cdot \frac{\cos\frac{\pi}{8}}{\cos\frac{\pi}{8}} = \frac{\sin\frac{\pi}{4}}{1+\cos\frac{\pi}{4}} = \frac{\sqrt{2}/2}{1+\sqrt{2}/2} = \sqrt{2} - 1$$
A: Your friend is saying  $$8(\arctan(\sqrt2-1)) = \pi$$
in other words removing inverse function he is saying 
$$  (\sqrt2-1)=\tan \frac{\pi}{8}=t $$
that you can say using double angle formula for tan
$$\tan \pi/4 = \frac{2t}{1-t^2} $$
which is same thing as saying
$$ 1=\frac{2 (\sqrt2-1)}{1-(3-2\sqrt2)}$$
thats same thing as saying ... after cancelling $2$ in numerator & denominator:
$$ 1=\frac{\sqrt2-1}{\sqrt2-1}$$
So he is right.
