Show that $f(x)=\pi-2\arctan(\sqrt{x-1})$

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 $$ \frac{\pi}{2}-g(x)=\arctan\sqrt{x-1} $$