Think of $\tan (x)$ as the gradient of a line drawn at $x$ degrees to the horizontal.

Consider the following matrix, which I call $R(x)$:

${\displaystyle \left(\begin{array}{cc}\hfill 1\hfill & \hfill m\hfill \\ \hfill -m\hfill & \hfill 1\hfill \end{array}\right)}$

where $m=\tan (x)$

It sends a horizontal line $(10)$ to $(1m)$ which is a line of gradient $m$. And it does something similar to the vertical line $(01)$. In otherwords it is a rotation [and an enlargement, but you don't notice that on lines through the origin] taking the horizontal to gradient $m$.

Hence $R(x)R(y)$ takes the horizontal line to one at an angle of $(x+y)$ to the horizontal - ie. one with gradient $\tan (x+y)$.

But $R(x).R(y)$ is $(m=\tan (x),n=\tan (y))$:

${\displaystyle \left(\begin{array}{cc}\hfill 1\hfill & \hfill m\hfill \\ \hfill -m\hfill & \hfill 1\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1\hfill & \hfill n\hfill \\ \hfill -n\hfill & \hfill 1\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1-mn\hfill & \hfill n+m\hfill \\ \hfill -n-m\hfill & \hfill 1-mn\hfill \end{array}\right)}$

So $(10)$ goes to $(1-mnn+m)$ which has gradient:

$(n+m)/(1-mn)$.

Hence $\tan (x+y)=(\tan (x)+\tan (y))/(1-\tan (x)\tan (y))$

Hope this is okay,