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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.

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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

4 Answers 4

up vote 11 down vote accepted

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 $$\tan(A+B) = \dfrac{\sin(A) \cos(B) + \cos(A) \sin(B)}{\cos(A) \cos(B) - \sin(A) \sin(B)} = \dfrac{\tan(A) + \tan(B)}{1 - \tan(A) \tan(B)}$$

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

But that is not what this answer is about.

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 $[0,1)$.

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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.

$f'(x) = g'(x)$ $f(x) = g(x) + \frac{\pi}{4}$

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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. –  B. 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.

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