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

Question

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 !

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

Answer 2

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.

Answer 3

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