trigonometry - Proving $tan(theta_1+theta_2+cdots+theta_n)$ has the form $frac{S_1-S_3+S_5+cdots}{1-S_2+S_4+cdots}$

$$\tan(\theta_1+\theta_2+\cdots+\theta_n)=\frac{S_1-S_3+S_5+\cdots}{1-S_2+S_4+\cdots}$$

where $S_i$ denotes the sum of product of tangent of angles taken $i$ at a time.

For example, $$\tan(\theta_1+\theta_2)=\frac{S_1}{1-S_2}=\frac{\tan\theta_1+\tan\theta_2} {1-\tan\theta_1\tan\theta_2}$$

(This formula is given in my textbook with no derivation or background)

How to derive this?

Answer

Prove that is true for $2$ angles, then consider it true for $(n-1)$ angles and prove for $n$ angles.
If you prove it for $2$ angles, it will be easy to prove it for $n$ angles (knowing it is true for $(n-1)$ angles) by considering $\theta_1+...+\theta_{n-1}$ is one angle, and $\theta_n$ is the other.