Find $x$ if $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$ 
The relation $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$ holds true for all $1.$ $ x\in \mathbb R$, $2.$ $ x\in (-1,\infty) $, $3.$ $ x\in (-\infty,1) $, $4.$ $ x\in (-\infty,2)$

I took RHS=$\frac{\pi}{4}+\tan^{-1}x=\tan^{-1}1+\tan^{-1}x=\tan^{-1}(\frac{1+x}{1-x})$,if $x\gt0, x\lt1$
(using the property $\tan^{-1}x+\tan^{-1}y=\tan^{-1}(\frac{x+y}{1-xy})$, if $x\gt0,y\gt0,xy\lt1$)
Therefore, my answer is $x\in(0,1)$, which is not an option.
The answer in the key is $2. $ $x\in(-\infty,1)$.
Looks like, they just used $x\lt1$ condition and not $x\gt0$. Why?
 A: Note that the function
$$f(x)= \tan^{-1}\frac{1+x}{1-x}-(\frac{\pi}{4}+\tan^{-1}x)$$
is continuous everywhere except at the break point $x=1$. Then, evaluate
$$\lim_{x\to 1^-} f(x)=0,\>\>\>\>\>\>\>
\lim_{x\to 1^+}f(x)=-\pi\ne 0
$$
Thus, the equality holds over $(-\infty,1)$.
A: A drawing might help.

Construct a right-angle triangle $\triangle ABC$ with sides $\overline{AB} = 1$ and $\overline{CB} = |x|$. Then
$$\measuredangle BAC =\arctan |x|.$$
Produce $AB$ to $D$ so that $\overline{BD} = \frac{|x|}{|y|}$. As a consequence you have
$$\measuredangle ADC = \arctan |y|.$$
By Euclid's Theorem, if $x^2 < \frac{|x|}{|y|}$, i.e. $|x|<\frac1{|y|}$, the angle $\angle ACD$ is obtuse. Draw from $D$ the line perpendicular to $AC$ and let $H$ be the foot of the altitude.
Similarity $\triangle ABC \sim \triangle ADH$ yield $$\overline{DH} = \frac{|x|(|x|+|y|)}{|y|\sqrt{1+x^2}},$$ and $$\overline{CH} = \frac{|x|(1-|xy|)}{|y|\sqrt{1+x^2}}$$.
External angle theorem gives
$$\angle HCD= \arctan |x| + \arctan |y|,$$
i.e.
$$\arctan\frac{|x|+|y|}{1-|xy|}=\arctan |x| + \arctan |y|,$$
valid for $|xy|<1$.
Thus we have, using $\arctan$'s odd simmetry,

*

*If $x>0$ and $0<y<\frac1x$, or if $x<0$ and $\frac1x < y<0$, then $$\arctan\frac{x+y}{1-xy}=\arctan x + \arctan y$$

*If $x>0$ and $-\frac1x<y<0$, or if $x<0$, then and $0<y<-\frac1x$ $$\arctan\frac{x-y}{1+xy}=\arctan x - \arctan y$$

You can analyze the situation when $|xy|>1$ in the same manner, considering that now the angle $\angle ACD$ is acute.
A: Taking the tangent of the two members, you check that this equation holds to a multiple of $\pi$.
The function $\arctan$ maps the real axis to the range $\left(-\dfrac\pi4,\dfrac\pi4\right)$ and is odd. Then $\dfrac\pi4+\arctan x\in\left(0,\dfrac\pi2\right)$ and this excludes the case of $x\ge1$.
