How to find the value of $\arctan(\frac{1}{1-x}) + \arctan(1-x)$?

Question

I'm reading a book on complex analysis. In one step while evaluating a path integral, the author makes the following substitution:

$$\arctan \left(\dfrac{1}{1-α} \right) + \arctan(1-α) = \dfrac{\pi}{2}$$

I'm not sure if it's relevant here, but $α$ is a complex number such that the absolute value of the real and imaginary parts are both less than $1$.

How did the author make this substitution? Is he using some sort of trigonometric identity I'm not aware of? If so, which one is it? And is there a list of such identities? I'd like to make sure I'm not confused next time I see something like this.

Related : http://math.stackexchange.com/questions/326334/a-question-about-the-arctangent-addition-formula and http://math.stackexchange.com/questions/587865/arctan-fracxy1xy-arctan-x-arctan-y-for-this-to-be-true-xy-has?lq=1 — lab bhattacharjee, Jan 14 '14 at 05:55
ArcTan[a] + ArcTan[b] = ArcTan[(a + b) / (1 - a b)] and, if a b =1, then .... — Claude Leibovici, Jan 14 '14 at 06:28
http://math.stackexchange.com/questions/610261/simplifying-an-arctan-equation — lab bhattacharjee, Jan 14 '14 at 07:59

score 5 · Answer 1 · answered Jan 14 '14 at 05:45

5

If $\tan \theta = u$, then $\cot \theta = \tan (\tfrac{\pi}{2} - \theta) = \frac{1}{u}$. The "co-" in the cofunctions refers to complementary angles.

answered Jan 14 '14 at 05:45

Sammy Black

25,273

score 3 · Answer 2 · answered Jan 14 '14 at 07:55

How about a triangle trigonometry answer? The tangent of an angle in a right triangle is the length of the opposite side divided by the length of the adjacent side. If you choose the other acute angle of the right triangle instead, the opposite and adjacent sides reverse themselves, so the tangent of the second angle is the reciprocal of the tangent of the first. Since both acute angles sum to $\frac\pi2$ radians, it makes sense that

$$\tan^{-1}a+\tan^{-1}\frac1a=\frac\pi2$$

score 2 · Accepted Answer · answered Jan 14 '14 at 05:45

2

Using $$\tan(a+b) = \frac{\tan a + \tan b}{1- \tan a \tan b}$$ Set $a$ equal to the first $\arctan$ and $b$ to the second $\arctan$. Clearly $ab =1$ so the two angles add to $\pm \pi/2$.

Sum could be $-\pi/2$ also, if the individual terms are negative (note $-\pi/2$ is also $3 \pi/2$.

answered Jan 14 '14 at 05:45

user44197

9,730

saw your comment about $\alpha$. In this case the final answer has to be $+\pi/2$. In general it could be either $\pm \pi/2$ – user44197 Jan 14 '14 at 05:47
Of course! At first I was confused becasue $tana tanb = 1$, which made the denominator $1 - 1 = 0$, which I thought was undefined. I forgot that $tan \frac{\pi}{2}$ itself is undefined. Thanks! – Brian Gesiak Jan 14 '14 at 06:09

score 1 · Answer 4 · answered Jan 14 '14 at 06:08

1

If you take the derivative of the function, you see it is constant. So plug in 0 for example to solve for the constant.

answered Jan 14 '14 at 06:08

Flowers

775
4
13

How to find the value of $\arctan(\frac{1}{1-x}) + \arctan(1-x)$?

4 Answers4