Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$ 
Given, for every $x>1$,
  $$f(x)=4\arctan\frac{1}{\sqrt{x-1}+\sqrt{x}}$$
  Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$

I have tried to use the fact that $\arctan(x)+\arctan(1/x)=\frac{\pi}{2
}$
So I obtain: $f(x)=4(\frac{\pi}{2}-\arctan(\sqrt{x-1}+\sqrt{x})$
I am stuck here !

Your statement is equivalent to proving that, for
$$
g(x)=2\arctan\frac{1}{\sqrt{x}+\sqrt{x-1}}
$$
we also have
$$
g(x)=\frac{\pi}{2}-\arctan\sqrt{x-1}
$$
Note that
$$
\frac{1}{\sqrt{x}+\sqrt{x-1}}=\sqrt{x}-\sqrt{x-1}
$$
Set $h(x)=\sqrt{x}-\sqrt{x-1}$ and prove that, for $x>1$, $0<h(x)<1$.
It follows that $0<g(x)<\pi/2$.
Now
\begin{align}
\tan g(x)
&=\tan(2\arctan h(x))\\[6px]
&=\frac{2\tan\arctan(h(x))}{1-\tan^2(\arctan h(x))}\\[6px]
&=\frac{2h(x)}{1-(h(x))^2}\\[6px]
&=2\frac{\sqrt{x}-\sqrt{x-1}}{1-x-(x-1)+2\sqrt{x(x-1)}}\\[6px]
&=\frac{\sqrt{x}-\sqrt{x-1}}{\sqrt{x-1}(\sqrt{x}-\sqrt{x-1})}\\[6px]
&=\frac{1}{\sqrt{x-1}}
\end{align}
Therefore
$$
\tan\left(\frac{\pi}{2}-g(x)\right)=
\cot g(x)=\sqrt{x-1}
$$
and so
$$
\frac{\pi}{2}-g(x)=\arctan\sqrt{x-1}
$$
