It is easily to see that $$ \arctan x + \arctan \frac {1} {x} = \begin {cases} \frac {\pi} {2}, \quad x>0, \ -\frac {\pi} {2}, \quad x<0, \end {cases} $$ and $$ \arctan z = -\arctan (-z),\quad \lim_ {z\to -\infty}\arctan z = -\frac {\pi} {2}, \quad \lim_ {z\to +\infty}\arctan z = \frac {\pi} {2}. $$ We divide the problem to three cases.
simple solution... draw a graph of the tangent function. Then draw the arctan function on both an x-y graph and a y-x graph.
Is there any formula for $$\operatorname {arctan} { (a+b)}$$ I know couple of formulas for trigonometric functions. But I don't know if such formulas exists for inverse trigonometric functions. I don'...
$\arctan(1/2)$ seems to be some strange, irrational angle, and the same goes for $\arctan(1/3)$, but those two angles seem to sum up to $45$ degrees. This seems like a mystery to me even though I can
Yes. $\cot x$ is the reciprocal, $\arctan x$ is the (principal) inverse, and $\arctan x=\frac1 {\tan x}$ is incorrect.
$ (1)$ If $\frac\pi2<\arctan x+\arctan y\le\pi, \arctan\left (\frac {x+y} {1-xy}\right)=\arctan x+\arctan y-\pi$ to keep $\arctan\left (\frac {x+y} {1-xy}\right)\in [-\frac\pi2,\frac\pi2]$
The notation $\tan^2 (x)$ "by default" denotes the square of the tangent because it is exceptional to meet the iterated function $\tan (\tan (x))$ that the exponent $2$ might denote. On the opposite, $\tan^ {-1} (x)$ might as well denote the functional inverse ($\arctan (x)$) or the multiplicative inverse ($\cot (x)$), because the two functions are frequently used. I guess that the first ...
For small angles, $\arctan x \approx x$. Most arctangents that you take will result in irrational numbers. In a programming environment, normally a floating point answer that is close enough to the truth is the best you will do. But in most of those environments, there is already a function available. So what are you trying to do?