# Operation on inverse trigonometric functions

Prove that:

$\tan^{-1}{6x-8x^3\over 1-12x^2} + \tan^{-1}{4x\over 1-4x^2} = \tan^{-1}2x$, $|2x|<{1\over \sqrt3}$

• Hint: take the tangent of both members. – Yves Daoust May 5 '16 at 10:49

Hint:

If $|\alpha\beta|<1$, then $$\arctan \alpha + \arctan \beta =\arctan\frac{\alpha+\beta}{1-\alpha\beta}$$

HINT:

Use the definition of Principal values,

As $-\dfrac\pi6\le\tan^{-1}(2x)\le\dfrac\pi6,$ $\tan^{-1}\dfrac{6x-8x^3}{1-12x^2}=\tan^{-1}\dfrac{3(2x)-(2x)^3}{1-3(2x)^2} =3\tan^{-1}(2x)$

As $-\dfrac\pi4\le\tan^{-1}(2x)\le\dfrac\pi4,$ $\tan^{-1}\dfrac{4x}{1-4x^2}= \tan^{-1}\dfrac{2(2x)}{1-(2x)^2}=2\tan^{-1}(2x)$

With some labor, you can prove more. Consider $$f(x)=\arctan\frac{6x-8x^3}{1-12x^2} \color{red}{-} \arctan\frac{4x}{1-4x^2}$$ which is defined for $x\notin\{1/2,-1/2,1/\sqrt{12},-1/\sqrt{12}\}$. Note the minus sign, otherwise the identity is false.

We can compute the derivative; first, the derivatives of the two fractions are, respectively, $$\frac{6(1+4x^2)^2}{(1-12x^2)^2}, \qquad \frac{4(1+4x^2)}{(1-4x^2)^2}$$ Moreover $$1+\left(\frac{6x-8x^3}{1-12x^2}\right)^2= \frac{1+12x^2+48x^4+64x^6}{(1-12x^2)^2}=\frac{(1+4x^2)^3}{(1-12x^2)^2}$$ and $$1+\left(\frac{4x}{1-4x^2}\right)^2=\frac{1+8x^2+16x^4}{(1-4x^2)^2}= \frac{(1+4x^2)^2}{(1-4x^2)^2}$$ so we have $$f'(x)=\frac{6}{1+4x^2}-\frac{4}{1+4x^2}=\frac{2}{1+4x^2}$$ which is the same as the derivative of $g(x)=\arctan(2x)$. Thus we can say that $$f(x)=\begin{cases} c_1+\arctan(2x) & x<-1/2 \\[4px] c_2+\arctan(2x) & -1/2<x<-1/\sqrt{12} \\[4px] c_3+\arctan(2x) & -1/\sqrt{12}<x<1/\sqrt{12} \\[4px] c_4+\arctan(2x) & 1/\sqrt{12}<x<1/2 \\[4px] c_5+\arctan(2x) & x>1/2 \end{cases}$$ Since you're interested in $c_3$, just note that $f(0)=0$, so $c_3=0$.

For completeness, we can compute $$c_1=\pi, \quad c_2=\pi/2, \quad c_3=0, \quad c_4=\pi \quad c_5=0$$

Here's a way to see that your formula is wrong:

bc 1.06
Copyright 1991-1994, 1997, 1998, 2000 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type warranty'.
x=1/(2*sqrt(12))
a((6*x-8*x^3)/(1-12*x^2))+a(4*x/(1-4*x^2))
1.40517450751406798853
a(2*x)
.28103490150281359770
a((6*x-8*x^3)/(1-12*x^2))-a(4*x/(1-4*x^2))
.28103490150281359773
`