Consider a function $f(x)= \arcsin (\frac {2x}{1+x^2}) + \arccos (\frac{1-x^2}{1+x^2}) +\arctan (\frac{2x}{1-x^2})-a\arctan x$ 
Consider a function $f(x)= \arcsin (\frac {2x}{1+x^2}) + \arccos (\frac{1-x^2}{1+x^2}) +\arctan (\frac{2x}{1-x^2})-a\arctan x$, where $a$ is any real constant. Find the value of $a$ if $f(x)=0$ for all x

Replacing $x$ with $\tan y$
$$\arcsin (\sin 2y) +\arccos (\cos 2y) +\arctan (\tan 2y)-a\arctan x=0$$
$$\implies 2y+2y+2y-ay=0$$
$$a=6$$
Alternatively, since $\cos$ is an even function
$$2y-2y+2y-ay=0$$
$$a=2$$
There is another value possible, according to the answer, which is $-2$. How do I obtain that?

For any $\;x\in\left]-\infty,-1\right[\;,\;$ it results that
$\arcsin\left(\dfrac{2x}{1+x^2}\right)=-\pi-2\arctan x\;,$
$\arccos\left(\dfrac{1-x^2}{1+x^2}\right)=-2\arctan x\;,$
$\arctan\left(\dfrac{2x}{1-x^2}\right)=\pi+2\arctan x\;.$
Hence, for all $\;x\in\left]-\infty,-1\right[,\;$ it results that
$\arcsin\left(\dfrac{2x}{1+x^2}\right)+\arccos\left(\dfrac{1-x^2}{1+x^2}\right)+ \arctan\left(\dfrac{2x}{1-x^2}\right)=$
$=-\pi-2\arctan x-2\arctan x+\pi+2\arctan x=$
$=-2\arctan x$
Consequently, $\;a=-2\;,\;$ for all $\;x\in\left]-\infty,-1\right[.$
