The tangent is often drawn on graphs as slope line. Given that the tangent is a function that returns us a single number, I don't understand where the slope line comes from? If it shows the essence of the tangent, good. But then what is the formula for this tangent line?

Why tangent show yoursel as slope line?
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It looks like you're being confused because you think of "tangent" as the name of a particular trigonometric function. But that is not the only meaning of that word.
In geometry, a "tangent" (or sometimes "tangent line") to a curve means a straight line that just touches the curve in question. That's exactly the line that's being drawn in this figure.
The trigonometric function is named for tangent lines: For angles between $0$ and $90^\circ$, $\tan(v)$ gives you the length of the segment of the tangent to the unit circle that lies between the "point of tangency" and the $x$-axis.
However, this naming doesn't mean that the tangent line has lost the right to be called a tangent. It had the name first!
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Think of it in this way, tangent or specifically $\tan \theta$ is defined as ratio of perpendicular to the base, which means, it is the rate at which a point moves upwards (perpendicular to $x$-axis) in comparison to what moves horizontally (along the $y$-axis). That is what slope means actually.
When you get ready to climb a mountain, and you try to get a sense of its slope, the same process goes on your back of mind, most of the time it is unnoticed. When you have to move upwards a greater distance in comparison to what you go in the forward direction, you say its slope is steeper.
In calculus, you will learn how we associate each point on a graph with a tangent (if you know, then we take the first-order differentiation of the curve at that point) and essentially it is also equal to the trigonometric tangent of the angle that the line makes with the positive direction of $x$-axis.
The slope of the tangent line on a circle centered at the origin is equal to the negative of $\tan(\theta)$, where theta is the normal angle between the positive $x$-axis and the line from the origin to the point $P$.
The equation of the line itself can be found from a standard point-slope form: $$y-y_0=m(x-x_0),$$ where $m$ is the slope, found using calculus, this relationship, or possibly other methods.