How do you calculate the arctan($0$)?

Hint: To solve this problem we should be aware of the fact that arctan($0$) is at which at which $\tan (x) = 0$ and x belongs to the range, $ - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}$(or $ - 90 < x < 90$ if you use degrees).
Also, $\tan (x) = \dfrac{{\sin x}}{{\cos x}}$.

Complete step by step solution:
We need to solve arctan($0$)
We know that, arctan($0$) is at which at which $\tan (x) = 0$.
Tangent of theta is defined as the ratio of sine of theta to cosine of theta, i.e., $\tan (x) = \dfrac{{\sin x}}{{\cos x}}$
So, we get that, $\tan (x) = 0$only when, $\sin x = 0$
We will use the below unit circle to generalize $\sin x = 0$.


We get that, $\sin x = 0$in the given range of x, $ - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}$
$x = 0 + n.2.\pi $or else,
$x = \pi + 2.\pi .n$
In the given case, $x = n.\pi $, where n is an integer.
Only x = 0 satisfies, $ - \dfrac{\pi }{2} < x < \dfrac{\pi }{2}$.
We get, $\sin x = 0$
$\tan (x) = 0$and,
arctan($0$) $ = 0$

Note:
Arctan($0$) is at which $\tan (x) = 0$. Tangent of theta is defined as the ratio of sine of theta to cosine of theta, i.e., $\tan (x) = \dfrac{{\sin x}}{{\cos x}}$. Thus, $\tan (x) = 0$ only when, $\sin x = 0$.