Difference of equal derivative functions value $f_1(x) = \text{tan}^{-1}(x)$
$f_2(x) = \text{tan}^{-1}\big(\frac{x-1}{x+1}\big)$
Where
$f_1^\prime(x) = f_2^\prime(x)=\frac{1}{x^2+1}$
Therefore
$f_1(x) - f_2(x) = \theta$
How to find values of $\theta$ mathematically. I have solved it graphically and the answers are $\frac{\pi}{4}$ and $-\frac{3\pi}{4}$ in my opinion. Thanks in advance.

Hint:
Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
$$\arctan x+\arctan(-1)=\begin{cases} \arctan\frac{x-1}{1-(-1)x} &\mbox{if } x\cdot(-1)<1\\ \pi+\arctan\frac{x-1}{1-(-1)x} & \mbox{if } x\cdot(-1)>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } x\cdot(-1)=1\end{cases} $$
