Question

What is the domain of $\arctan \left( x \right)$?
(a) real numbers
(b) natural numbers
(c) $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$
(d) None of these

Answer 1

In this given problem, we are given with the arctan function which is said to be the inverse function of the tangent function. So, to find the result, we first check out the range of the function $\tan \left( x \right)$. Once we have done that, we need to check if we can form a function between that range and the domain of $\tan \left( x \right)$, which is well defined. And if we can find something of that sort, we will get our result.

Complete step by step solution:
According to the question, we are trying to find the domain of $\arctan \left( x \right)$.
To start with, In mathematics, the inverse trigonometric functions (occasionally also called arcus functions, anti-trigonometric functions or cyclometric functions) are the inverse functions of the trigonometric functions (with suitably restricted domains). Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios.
Thus, $\arctan \left( x \right)$tells us about the inverse function of the tangent function.
The function tan(x) is a many to one periodic function, so to define an inverse function requires that we restrict its domain (or restrict the range of the inverse function).
So, the inverse function gives us a domain of the inverse function of the tangent function.
The function $\tan \left( x \right)$ is a many to one periodic function, so to define an inverse function requires that we restrict its domain (or restrict the range of the inverse function). Now again, if we consider the domain of $\tan \left( x \right)$ to $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ , the function is one to one, continuous and unbounded over the given interval. Thus, we get a well defined function from the real numbers to $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$as $\arctan \left( x \right)$.
Hence, the domain of $\arctan \left( x \right)$is the whole of $\mathbb{R}$that is $\left( -\infty ,\infty \right)$ .
So, we have our solution as, (a) real numbers.

Note: In this problem, we have used the function $\arctan \left( x \right)$to get our solution. There are some specific properties of $\arctan \left( x \right)$to deal with. Some of them are stated here:
1.$\tan \left( \arctan \left( x \right) \right)=x$ 2.$\arctan \left( -x \right)=-\arctan \left( x \right)$ 3.$\arctan \left( a \right)\pm \arctan \left( b \right)=\arctan \left( \dfrac{a\pm b}{1\mp ab} \right)$