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?

  • $\begingroup$ The functions like $\sin^{-1}\sin 2y$ in both of your equations are periodic, their periodicity must be taken into account $\endgroup$
    – Z Ahmed
    Oct 7, 2020 at 17:02
  • $\begingroup$ @ZAhmed I don’t know how to apply that information here. It would leave a residual $\pi$ $\endgroup$
    – Aditya
    Oct 7, 2020 at 17:10
  • $\begingroup$ Use en.wikipedia.org/wiki/… $\endgroup$ Oct 7, 2020 at 17:48

1 Answer 1


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 $\;x\in\left]-\infty,-1\right[.$


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