<s>[INST] 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 !
 [/INST] 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
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