On definition of Tangente: Why is $\tan(\theta + \pi) = \frac{-b}{-a}$?

Question

This is from IB Standard Level. Students I guess are not allowed to use formulas for $\sin(A+B), \tan(A+B)$, etc.

Students know $\sin(\theta)=b, \cos(\theta)=a, \tan=\sin/\cos, \cos^2+\sin^2=1, |\cos|, |\sin| \le 1$, arc length and area of sector formulas, equations of circles

Answer 1

A point $(x,y) \neq (0,0)$ determines an angle $\theta$ as measured counter-clockwise from the $x$-axis. If $x\neq0$, then we define $$\tan\theta = \frac yx.$$

Therefore, following your picture, we have $$\tan(\theta+\pi) = \frac{-y}{-x} = \frac yx = \tan\theta.$$

Answer 2

In fact the definition of $\tan \theta$ geometrically does not use the formula $$\tan \theta=\frac{\sin \theta }{\cos \theta} $$

BUT The Pure Defintion of $\tan $ is as follows:

Geometrically, If $P$ denote the intersection point of the line (L1) perpendicular to the $OI-$axis passing through $I(1,0)$ and the line $(OP)=(L_2(\theta))$ passing through $O(0,0)$ with inclination angle $\color{red}{\theta}$

Then Definition $$\color{red}{\tan\theta = \overline{IP}}~~~\text{where }~~I(1,0)$$ $\overline{IP}$ denote the algebraic distance from $I$ to $P$.

It easy is to see from the figure that lines L2($\theta$)) and L2($\pi +\theta$)) are identical i.e $$\color{red}{(L_2(\theta))\equiv (L_2(\pi +\theta)).}$$

therefore by defintion of $\tan $ given above

$$ \color{blue}{\frac{a}{b}=\tan \theta = \tan (\pi+ \theta) =\frac{-a}{-b}} $$