# 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?

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

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