# Prove that $\tan^{-1}\left(\frac{x+1}{1-x}\right)=\frac{\pi}{4}+\tan^{-1}(x)$

The question is:

Prove that $\tan^{-1}\left(\frac{x+1}{1-x}\right)=\frac{\pi}{4}+\tan^{-1}(x)$.

It's from A-level further mathematics.

-
Welcome to math.SE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, people are also much happier to help those who demonstrate that they've tried the problem themselves first. – Zev Chonoles Jun 20 '12 at 5:03
Please see here and here for how to format your mathematics expressions with LaTeX, and see here for how to use Markdown formatting. If you need to format more advanced math, there are many excellent LaTeX references on the internet, including Stack Exchange's own TeX.SE site. If you see a piece of LaTeX you want to know the code for on the site, you can right click on it, go to "Show Math As", then choose "TeX Commands". – Zev Chonoles Jun 20 '12 at 5:04
Given Problem is incorrect. There should be (1-x) in place of (x-1). You can check it simply by putting x=0(relation doesn't hold). – Aang Jun 20 '12 at 5:22
@E.O.: Marvis got the valid relation.problem is with the "problem". – Aang Jun 20 '12 at 5:24
Initially, I had the wrong sign as well. – user17762 Jun 20 '12 at 5:25

The identity should read $$\tan^{-1} \left(\dfrac{x+1}{1-x} \right) = \tan^{-1}(x) + \pi/4$$

Let $\tan^{-1}(x) = \theta$ i.e. $x = \tan(\theta)$. Then we get that $$\dfrac{x+1}{1-x} = \dfrac{\tan(\theta) + 1}{1-\tan(\theta)}$$ Recall that $\tan(A+B) = \dfrac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$.

Taking $B = \pi/4$, we get that $\tan(A+ \pi/4) = \dfrac{\tan(A) + 1}{1 - \tan(A)}$.

Hence, we get that $$\dfrac{x+1}{1-x} = \dfrac{\tan(\theta) + 1}{1-\tan(\theta)} = \tan(\theta + \pi/4)$$ Hence, $$\tan^{-1} \left(\dfrac{x+1}{1-x} \right) = \theta + \pi/4 = \tan^{-1}(x) + \pi/4$$

Proof of $\tan(A+B) = \dfrac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$

$$\tan(A+B) = \dfrac{\sin(A+B)}{\cos(A+B)} = \dfrac{\sin(A) \cos(B) + \cos(A) \sin(B)}{\cos(A) \cos(B) - \sin(A) \sin(B)}$$ Assuming $\cos(A) \cos(B) \neq 0$, divide numerator and denominator by $\cos(A) \cos(B)$, to get that $$\tan(A+B) = \dfrac{\sin(A) \cos(B) + \cos(A) \sin(B)}{\cos(A) \cos(B) - \sin(A) \sin(B)} = \dfrac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$$

-
@E.O. Yes edited accordingly. Thanks. – user17762 Jun 20 '12 at 5:24

Apply the function tan to both sides. Using the addition law for tan, we find that $$\tan\left(\frac{\pi}{4}+\tan^{-1}x\right)=\frac{1+x}{1-x}.$$

Note that in the usual definition of $\tan^{-1} x$, we say that $\tan^{-1} x$ is the number (angle) in a specific interval whose tangent is $x$. For example, the Wikipedia definition specifies that $\tan^{-1} x$ is the number between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ whose tangent is $x$.

Under that definition, the supposed identity is not correct. Take for example $x=\sqrt{3}$. Then $\tan^{-1} x= \frac{\pi}{3}$. So the right-hand side is greater than $\frac{\pi}{2}$, while the left-hand side is negative.

Similar considerations hold if we for example specify that $\tan^{-1}x$ lies between $0$ and $\pi$. We can make the identity correct by restricting $x$ to the interval $[0,1)$.

-

Although Marvis's answer is more complete (+1) that you had expected, I am adding other. We know that If $f'(x) = g'(x)$, then $f(x) = g(x) + C$ for some constant $C$.. Here, if you put $f(x)=\tan^{-1}(\frac{x+1}{1-x})$ and $g(x)=\tan^{-1}(x)$; then $f'(x) = g'(x)$. So there is a constant $C$ that $f(x) = g(x) + C$. Now, put $x=0$ into both sides of the latter equality.

-
What are the graphs? – Gerry Myerson Jun 20 '12 at 7:36
@GerryMyerson: The first one shows that $f'(x)$ and $g'(x)$ have the same graphs and the second one shows that graphs of $\tan^{-1}(\frac{x+1}{1-x})$ and $\frac{\pi}{4}+\tan^{-1}(x)$ when $x<1$,coincides. – Babak S. Jun 20 '12 at 7:59
pretty graphics, again! + 1 – amWhy Mar 6 '13 at 0:56

I'll resort to 2 very powerful formulas i did at school,often used, namely $\cos(2\tan^{-1}(x))=\frac{1-x^2}{1+x^2}$ and $\sin(2\tan^{-1}(x))=\frac{2x}{1+x^2}$. After multiplying the both sides of the identity by 2 and taking cos
of them, we simply get that: $$\cos\left(2\tan^{-1}\left(\frac{x+1}{1-x}\right)\right)=\cos\left(\frac{\pi}{2}+2\tan^{-1}(x)\right) \tag1$$

The calculations to do here are very simple and immediately get that:

$$\frac{-2x}{x^2+1}=\frac{-2x}{x^2+1}.$$ Q.E.D.

-