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 !
 A: 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
$$
\frac{\pi}{2}-g(x)=\arctan\sqrt{x-1}
$$
A: Hint: Differentiate both expressions. If the derivative is the same then the functions can only be differring by a constant. Plug in $x=1$ to determine the constant. If both expressions are the same the constant should be 0.
EDIT: Thanks to @Olivier Oloa, it seems that both expressions are not the same or you have a typo.
A: Observe that
$$\frac1{\sqrt{x-1}+\sqrt x}=\sqrt x-\sqrt{x-1}$$
and thus
$$\left(4\arctan(\sqrt x-\sqrt{x-1})\right)'=4\left(\frac1{2\sqrt x}-\frac1{2\sqrt{x-1}}\right)\cdot\frac1{1+(\sqrt x-\sqrt{x-1})^2}=$$
$$=2\left(\frac1{\sqrt x}-\frac1{\sqrt{x-1}}\right)\frac1{2\sqrt x(\sqrt x-\sqrt{x-1})}=\frac{\sqrt{x-1}-\sqrt x}{x\sqrt{x-1}(\sqrt x-\sqrt{x-1})}=$$
$$=\color{red}{-\frac1{x\sqrt{x-1}}}$$
And on the other hand:
$$\left(\pi-2\arctan\sqrt{x-1}\right)'=-\frac1{\sqrt{x-1}}\frac1{1+x-1}=\color{red}{-\frac1{x\sqrt{x-1}}}$$
Thus, both forms of the $\;f\;$ have the same derivative and thus they differ only by a constant, say $\;K\;$ :
$$\pi-2\arctan\sqrt{x-1}=4\arctan(\sqrt x-\sqrt{x-1})+K$$
and observe that in the above form both sides are well defined for $\;x=1\;$, so substituting $\;x=1\;$ :
$$\pi=\pi-2\arctan0=4\arctan(1)+K=4\frac\pi4+K=\pi+K\implies K=0$$
and we get the wanted equality
A: Using trigonometry is much esasier and more revealing about the geometrical meaning of this type of identities.
Consider the Figure below.

The right-angled triangle $\triangle ABC$ has sides $\overline{AB} = 1$ and $\overline{BC} = \sqrt{x-1}$, so that
$$\alpha = \arctan \sqrt{x-1},$$
and, by Pythagorean Theorem, $\overline{AC} = \sqrt x$.
Now extend $BC$ to a segment $\overline{CD} = \overline{AC} = \sqrt x$,
so that
$$\beta = \arctan \frac{1}{\sqrt{x-1} + \sqrt x}.$$
Use finally the fact that $\triangle ACD$ is isosceles and that $\triangle ABD$ is right-angled to write
$$ 2\beta + \alpha = \frac{\pi}{2},$$
which is exactly equivalent to your identity, once you plug in the definitions of $\alpha$ and $\beta$. As simple as that. $\blacksquare$
