Let us assume that function $g$ is defined on an interval $(a,b)$, $g(x)\in(a,b)$ in that interval, and that there is a constant $c<1$ such that for each $x,y \in (a,b)$,
\begin{equation}
\left|g(y)-g(x)\right| < c |x-y|. \tag{1}
\end{equation}
If $g$ has a derivative, this becomes $g'(x)<c$ for $x\in(a,b)$.
The fixed point iteration is defined by $x_{k+1}=g(x_k)$, where $x_0$ is an arbitrarily chosen starting point in $(a,b)$.
Let us assume that the function has a fixed point at $\hat{x}\in(a,b)$, that is $\hat{x}=g(\hat{x})$.
Now at step $k$, the absolute error of our current guess to the fixed point is $e_k = |x_k-\hat{x}|$. We get
$$
e_{k+1} = |x_{k+1}-\hat{x}| = |g(x_k)-g(\hat{x})| < c|x_k - \hat{x}| = c e_k.
$$
Therefore, the sequence $(e_k)_k$ is nonnegative and bounded above by the sequence $(c^ke_0)_k$, which converges to $0$. Therefore, $\lim_{k\to\infty}e_k=0$. This means that the fixed point iteration converges to $\hat{x}$.
For general $g:\mathbb{R}\to\mathbb{R}$, we can make following observations:
If (1) holds in $\mathbb{R}$, we can replace $(a,b)$ with $\mathbb{R}$ in the above proof. One can also see that the function has exactly one fixed point in that case (if $g$ is differentiable, the derivative of $g(x)-x$ is smaller than a negative constant, thus $g(x)-x$ has exactly one zero; if $g$ is not differentiable, a similar argument still holds).
If (1) does not hold in $\mathbb{R}$ but holds in an interval $(a,b)$ containing a fixed point, we can see that $g(a)>a$ and $g(b)<b$, so $g(x) \in (a,b)$ as required. Now the fixed point iteration converges to the fixed point if $x_0$ is chosen inside the interval.
The $\arctan$ function is more nicely behaved than $\tan$. Let
$$g(x)=\arctan x+\pi.$$
Be sure your calculator is in radian mode, and start with $x_0$ not too far from $4.5$. Use the iteration $x_{n+1}=g(x_n)$. Convergence should be acceptably quick.
Added: The reason we get nice behaviour with our choice of $g(x)$ is that near the root, the derivative of $\arctan x$ has rather small absolute value. By way of contrast, the derivative of $\tan x$ near the root is much larger than $1$. That means that with the choice $g(x)=\tan x$, the root is a repelling fixed point.
Best Answer
What definition of trig functions (also called "circular functions") are you using?
The definitions in terms of right angles cannot be used in general as that requires the angle to be between 0 and 90 degrees but, in order to use them as functions we want them defined for all real numbers.
Of the "circle" definition is used: Construct a unit circle (center at (0, 0), radius 1) in an xy-coordinate system. Given non-negative number, t, measure, counter clockwise, around the circumference a distance t (for negative t, measure clockwise). The endpoint has coordinates $(\cos(t), \sin(t))$. That is, whatever the end point is, the x coordinate is, by definition, $\cos(t)$, the y coordinate is $\sin(t)$.
Notice that "$t$" is NOT in degrees because it is not an angle at all, it is a distance around the circumference. We can think of that as an angle by identifying the angle with the distance subtended on a unit circle which is simply the definition of "radian".