Find $x$ if $\tan^{-1}(\frac{1+x}{1-x})=\frac{\pi}{4}+\tan^{-1}x$

Question

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?

Answer 1

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,

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

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You can analyze the situation when $|xy|>1$ in the same manner, considering that now the angle $\angle ACD$ is acute.

Answer 2

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 $(-\infty,1)$.

Answer 3

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 $x\ge1$.